# Magnetic Field due to a Current

## Concept Of Physics

### H C Verma

1   1- Using the formulae $\vec{F} = q\vec{v} \times \vec{B}$ and B = $\frac{\mu_0 I}{2\pi r}$, show that the SI units of the magnetic field B and the permeability constant $\mu_0$ may be written as N/A-m and $N/A^2$ respectively.

##### Solution :

$\vec{F} = q\vec{v} \times \vec{B}$ or, $B = \frac{F}{qv} = \frac{F}{iTv} = \frac{N}{A.sec./sec.} = \frac{N}{A - m}$$\\$$B = \frac{\mu_0 I}{2\pi R}$ or, $\mu_0 = \frac{2\pi rB}{I} = \frac{m\times N}{A - m\times A} = \frac{N}{A^2}$

2   2- A current of 10 A is established in a long wire along the positive Z-axis. Find the magnetic field $\vec{B}$ at the point (1 m, 0, 0).

##### Solution :

i = 10A, d = 1m$\\$$B = \frac{\mu_0 I}{2\pi r} = \frac{10^{-7}\times 4\pi \times 10}{2\pi \times 1} = 20\times 10^{-6} T = 2\mu T$$\\$Along +ve Y direction.

3   3- A copper wire of diameter 1.6 mm carries a current of 20 A. Find the maximum magnitude of the magnetic field $\vec{B}$ due to this current.

d = 1.6 mm$\\$ So, r = 0.8 mm = 0.0008 m$\\$ i = 20 A$\\$$\vec{B} = \frac{mu_0 i}{2\pi r} = \frac{4\pi \times 10^{-7} \times 20}{2\times \pi 8\times 10{-4}} = 5\times 10^{-3} T = 5mT 4 4- A transmission wire carries a current of 100 A. What would be the magnetic field B at a point on the road if the wire is 8 m above the road? ##### Solution : i = 100 A, d = 8 m\\$$B = \frac{\mu_0 i}{2\pi r}$$\\$$= \frac{4\pi\times 10^{-7}\times 100}{2\times \pi \times 8} = 2.5\mu T$

5   5- A long, straight wire carrying a current of 1.0 A is placed horizontally in a uniform magnetic field B = 1.0 x $10^{-5}$ T pointing vertically upward (figure 35-E1). Find the magnitude of the resultant magnetic field at the points P and Q, both situated at a distance of 2.0 cm from the wire in the same horizontal plane.

$\mu_0 = 4\pi \times 10^{-7} T-m/A$$\\$$r = 2 cm = 0.02 m, I = 1 A, \vec{B} = 1\times 10^{-5} T$$\\ We know: Magnetic field due to a long straight wire carrying current = \frac{\mu_0 I}{2\pi r}$$\\$$\vec{B} at P = \frac{4\pi \times 10^{-7}\times 1}{2\pi \times 0.02} = 1\times 10^{-5} T upward\\net B = 2 × 1 × 10^{-7} T = 20 \muT\\B at Q = 1 × 10^{-5} T downwards\\Hence net \vec{B} = 0 6 6- A long, straight wire of radius r carries a current i and is placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross-section, (a) At what points will the resultant magnetic field have maximum magnitude ? What will be the maximum magnitude ? (b) What will be the minimum magnitude of the resultant magnetic field ? ##### Solution : (a) The maximum magnetic field is B+\frac{\mu_0 I}{2\pi r} which are along the left keeping the sense along the direction of traveling current.\\(b)The minimum B - \frac{\mu_0 I}{2\pi r}$$\\$If $r = \frac{\mu_0 I}{2\pi B}$ B net = 0$\\$$r < \frac{\mu_0 I}{2\pi B} B net = 0\\$$r > \frac{\mu_0 I}{2\pi B}$ B net = B - $\frac{\mu_0 I}{2\pi r}$

7   7- A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of $4.0\times 10^{-4}$ parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire.

$\mu_0 = 4\pi \times 10^{-7}$ T-m/A, I = 30 A, $B = 4.0\times 10^{-4}T$ Parallel to current.$\\$$\vec{B} due to wore at a pt. 2 cm$$\\$ $= \frac{\mu_0I}{2\pi r}$ = $\frac{4\pi \times 10^{-7}\times 30}{2\pi \times 0.02} = 3\times 10^{-4} T$$\\net field = \sqrt{(3\times 10^{-4})^2 + (4\times 10^{-4})^2} = 5\times 10^{-4} T 8 8- A long, vertical wire carrying a current of 10 A in the upward direction is placed in a region where a horizontal magnetic field of magnitude 2.0\times 10^{-3} T exists from south to north. Find the point where the resultant magnetic field is zero. ##### Solution : i = 10 A. \hat{(K)}$$\\$$B = 2 \times 10^{-3}T South to North \hat{(J)}$$\\$To cancel the magnetic field the point should be choosen so that the net magnetic field is along - $\hat{J}$ direction.$\\$$\therefore The point is along -\hat{i} direction or along west of the wire.\\$$B = \frac{\mu_0 I}{2\pi r}$$\\$$\Rightarrow 2\times 10^{-3} = \frac{4\pi \times 10{-7}\times 10}{2\pi \times r}$$\\$$\Rightarrow r = \frac{2\times 10^{-7}}{2\times 10^{-3}} = 10^{-3} m = 1 mm$

9   9- Figure (35-E2) shows two parallel wires separated by a distance of 4.0 cm and carrying equal currents of 10 A along opposite directions. Find the magnitude of the magnetic field B at the points $A_1$, $A_2$ , $A_3$ and $A_4$ .

Let the tow wires be positioned at O & P$\\$ R = OA, = $\sqrt{(0.02)^2 + (0.02)^2} = \sqrt{8\times10^{-4}} = 2.828\times 10^{-2} m$$\\(a) \vec{B} due to Q, at A_1 = \frac{4\pi \times 10^{-7}\times 10}{2\pi \times 0.02} = 1\times 10^{-4} T (\perpr towards up the line)\\$$\vec{B}$ due to P, $A_1$ = $\frac{4\pi \times 10^{-7}\times 10}{2\pi \times 0.06} = 0.33\times 10^{-4}$ ($\perp$r towards down the line)$\\$$\\net \vec{B} = 1\times 10^{-4} - 0.33\times 10^{-4} T$$\\$(b) $\vec{B}$ due to O at $A_2$ = $\frac{2\times10^{-7}\times 10}{0.01} = 2\times 10^{-4} T$ ($\perp$r down the line)$\\$$\vec{B} due to P at A_2 = \frac{2\times 10^{-7}\times 10}{0.03} = 0.67\times 10^{-4} \perpr towards down the line \\Net \vec{B} at A_2 = 2\times 10^{-4} + 0.67\times 10^{-4} = 2.67\times 10^{-4} T$$\\$(c) $\vec{B}$ at $A_3$ due to O = $1\times 10^{-4} T$ $\perp$r towards down the line $\\$$\vec{B} at A_3 due to P = 1\times 10^{-4} T \perpr towards down the line\\Net \vec{B} at A_3 = 2\times 10^{-4} T$$\\$(d) $\vec{B}$ at $A_4$ due to O = $\dfrac{2\times 10^{-7}\times 10}{2.828\times 10^{-2}} = 0.7\times 10^{-4}$ T towards SE$\\$$\vec{B} at A_4 due to P = 0.7\times 10^{-4} T towards SW \\ Net \vec{B} = \sqrt{(0.7\times 10^{-4})^2 + (0.7\times 10^{-4})^2} = 0.989\times 10^{-4} = 1\times 10^{-4} T 10 10- Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is 2.0 cm away from each of these wires. ##### Solution : Cos \theta = \frac{1}{2}, \theta = 60^{\circ} , and \angle AOB = 60^{\circ}$$\\$$B = \frac{\mu_0 I}{2\pi r} = \frac{10^{-7}\times 2\times 10}{2\times 10^{-2}} = 10^{-4}T$$\\$So net is $[(10^{-4})^2 + (10^{-4})^2 + 2(10^{-8}) Cos 60°]^\frac{1}{2}$$\\=10^{-4}[1 + 1 + 2\times \frac{1}{2}]^\frac{1}{2} = 10^{-4} \times \sqrt {T} = 1.732 \times 10^{-4} T$$\\$

11   11- Two long, straight wires, each carrying a current of 5 A, are placed along the X- and Y-axes respectively. The currents point along the positive directions of the axes. Find the magnetic fields at the points (a) (1 m, 1 m), (b) ( - 1 m, 1 m), (c) ( - 1 m, - 1 m) and (d) ( 1 m, - 1 m).

##### Solution :

(a) $\vec{B}$ for X = $\vec{B}$ for Y$\\$Both are oppositely directed hence net $\vec{B} = 0$$\\(b) \vec{B} due to X = \vec{B} due to X both directed along Z–axis\\Net \vec{B} = \frac{2\times 10^{-7}\times 2\times 5}{1} = 2\times 10^{-6} T = 2\mu T$$\\$(c) $\vec{B}$ due to X = $\vec{B}$ due to Y both directed opposite to each other. Hence Net $\vec{B}$ = 0$\\$(d) $\vec{B}$ due to X = $\vec{B}$ due to Y = $1\times 10^{-6} T$ both directed along (–) ve Z–axis. Hence Net $\vec{B}$ = $2\times 1.0\times 10^{-6} = 2 \mu T$

12   12- Four long, straight wires, each carrying a current of 5.0 A, are placed in a plane as shown in figure (35-E3). The points of intersection form a square of side 5.0 cm. (a) Find the magnetic field at the centre P of the square. (b) $Q_1 Q_2 , Q_3$ and $Q_4$ are points situated on the diagonals of the square and at a distance from P that is equal to the length of the diagonal of the square. Find the magnetic fields at these points.

##### Solution :

(a) For each of the wire Magnitude of magnetic field$\\$ $= \dfrac{\mu_0 i}{4\pi r}(sin 45^{\circ} + sin 45^{\circ}) = \dfrac{\mu_0\times 5}{4\pi \times (\dfrac{5}{2})}$$\dfrac{2}{\sqrt 2} \\For AB {\odot} for BC \odot For CD \otimes and for DA \otimes.\\The two odot and 2otimes fields cancel each other. Thus B_net = 0\\(b) At point Q_1$$\\$due to (1) B = $\dfrac{\mu_0 i}{2\pi \times 2.5\times 10^{-2}} = \dfrac{4\pi \times 5\times 2\times 10^{-7}}{2\pi \times 5\times 10^{-2}} = 4\times 10^{-5} \odot$$\\due to (2) B = \dfrac{\mu_0 i}{2\pi \times (\frac{15}{2})\times 10^{-2}} = \dfrac{4\pi \times 5\times 2\times 10^{-7}}{2\pi \times 15\times 10^{-2}} = (\dfrac{4}{3})\times 10^{-5} \odot$$\\$due to (3) B = $\dfrac{\mu_0 i}{2\pi \times (5+\dfrac{5}{2})\times 10^{-2}} = \dfrac{4\pi \times 5\times 2\times 10^{-7}}{2\pi \times 15\times 10^{-2}} = (\dfrac{4}{3})\times 10^{-5} \odot$$\\due to (4) B = \dfrac{\mu_0 i}{2\pi \times 2.5\times 10^{-2}} = \dfrac{4\pi \times 5\times 2\times 10^{-7}}{2\pi \times 5\times 10^{-2}} = 4\times 10^{-5} \odot$$\\$$B_{net} = [4 + 4 + (\frac{4}{3}) + (\frac{4}{3})]\times 10^{-5} = \frac{32}{3}\times 10^{-5} = 10.6\times 10^{-5} = 1.1\times 10^{-4} T$$\\$At point $Q_2$$\\due to (1) \dfrac{\mu_0i}{2\pi \times (2.5)\times 10^{-2}} \odot$$\\$due to (2) $\dfrac{\mu_0i}{2\pi \times (\dfrac{15}{2})\times 10^{-2}} \odot$$\\due to (3) \dfrac{\mu_0i}{2\pi \times (2.5)\times 10^{-2}} \otimes$$\\$due to (4) $\dfrac{\mu_0i}{2\pi \times (\dfrac{15}{2})\times 10^{-2}} \otimes$$\\$$B_{net}$ = 0$\\$At point $Q_3$$\\due to (1) = \dfrac{4\pi \times 10^{-7}\times 5}{2\pi \times (\dfrac{15}{2})\times 10^{-2}} = \dfrac{4}{3}\times 10^{-5}\otimes$$\\$due to (2) = $\dfrac{4\pi \times 10^{-7}\times 5}{2\pi \times (\dfrac{5}{2})\times 10^{-2}} = 4\times 10^{-5}\otimes$ $\\$

due to (3) = $\dfrac{4\pi \times 10^{-7}\times 5}{2\pi \times (\dfrac{5}{2})\times 10^{-2}} = 4\times 10^{-5}\otimes$$\\due to (4) = \dfrac{4\pi \times 10^{-7}\times 5}{2\pi \times (\dfrac{15}{2})\times 10^{-2}} = \dfrac{4}{3}\times 10^{-5}\otimes$$\\$$B_{net} =[4 + 4 + (\frac{4}{3}) + (\frac{4}{3})] \times 10^{-5} = \frac{32}{3} \times 10^{-5} = 10.6\times 10^{-5} = 1.1\times 10^{-4}T$$\\$For $Q_4$$\\due to (1) \frac{4}{3} \times 10^{-5} \otimes$$\\$due to (2) $4\times 10^{-5} \otimes$$\\due to (3) \frac{4}{3} \times 10^{-5} \otimes$$\\$due to (4) $4\times 10^{-5} \otimes$$\\$$B_{net}$ = 0

13   13- Figure (35-E4) shows a long wire bent at the middle to form a right angle. Show that the magnitudes of the magnetic fields at the points P, Q, R and S are equal and find this magnitude.

Since all the points lie along a circle with radius = ‘d’. Hence ‘R’ & ‘Q’ both at a distance ‘d’ from the wire. So, magnetic field $\vec{B}$ due to are same in magnitude. As the wires can be treated as semi infinite straight current carrying conductors. Hence magnetic field $\vec{B} = \frac{\pi_0 i}{4\pi d}$$\\At P\\$$B_1$ due to 1 is 0$\\$$B_2 due to 2 is \frac{\pi_0 i}{4\pi d}$$\\$At Q $\\$$B_1 due to 1 is\frac{\pi_0 i}{4\pi d}$$\\$$B_2 due to 2 is 0\\At R\\$$B_1$ due to 1 is 0$\\$$B_2 due to 2 is \frac{\pi_0 i}{4\pi d}$$\\$At S$\\$$B_1 due to 1 is \frac{\pi_0 i}{4\pi d}$$\\$$B_2 due to 2 is 0 14 14- Consider a straight piece of length x of a wire carrying a current i. Let P be a point on the perpendicular bisector of the piece, situated at a distance d from its middle point. Show that for d >> x, the magnetic field at P varies as \frac{1}{d^2} whereas for d << x, it varies as \frac{1}{d}. ##### Solution : B = \dfrac{\pi_0i}{4\pi d} 2 sin\theta$$\\$= $\dfrac{\pi_0 i}{4\pi d}$ $\dfrac{2\times x}{2\times \sqrt{d^2 + \dfrac{x^2}{4}}}$ = $\dfrac{\mu_0 i x}{4 \pi d \sqrt{d^2 + \dfrac{x^2}{4}}}$$\\(a) When d >> x\\Neglecting x w.r.t. d\\$$ B = \dfrac{\mu_0 ix}{\mu \pi d\sqrt{d^2}} = \dfrac{\mu_0 ix}{\mu \pi d^2}$$\\$$\therefore B \propto \dfrac{1}{d^2}$$\\(b) When x >> d, neglecting d w.r.t. x\\$$ B = \dfrac{\mu_0 ix}{4\pi \frac{dx}{2}} = \frac{2\mu_0 i}{4\pi d}$$\\$$\therefore B \propto \dfrac {1}{d}$

15   15- Consider a 10 cm long piece of a wire which carries a current of 10 A. Find the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece.

##### Solution :

I = 10 A, a = 10 cm = 0.1 m$\\$r = OP = \dfrac{\sqrt 3}{2}\times 0.1 m$\\$$B = \dfrac{\mu_0 I}{4\pi r}(sin \phi_1 + sin \phi_2)$$\\$= $\dfrac{10^{-7}\times 10\times 1}{\dfrac{\sqrt 3}{2}\times 0.1}$ = $\dfrac{2\times 10^{-5}}{1.732} = 1.154\times 10^{-5} T = 11.54\mu T$

16   16- A long, straight wire carries a current i. Let $B_1$ be the magnetic field at a point P at a distance d from the wire. Consider a section of length I of this wire such that the point P lies on a perpendicular bisector of the sectionl Let $B_2$ be the magnetic field at this point due to this section only. Find the value of $\frac{d}{l}$ so that $B_2$ differs from $B_1$ by 1%.

$B_1 = \dfrac{\mu_0 i}{2\pi d}$ , $B_2 = \dfrac{\mu_0i}{4\pi d}(2\times sin \theta)$ $= \dfrac{\mu_0i}{4\pi d}\dfrac{2\times l}{2\sqrt{d^2 + \dfrac{l^2}{4}}}$ = $\dfrac{\mu_0 il}{4\pi d \sqrt{d^2 + \dfrac{l^2}{4}}}$$\\$$B_1 - B_2 = \dfrac{1}{100} B_2 \Rightarrow \dfrac{\mu_0 i}{2\pi d} - \dfrac{\mu_0 il}{4\pi d \sqrt{d^2 + \dfrac{l^2}{4}}} = \dfrac{\mu_0 i}{200\pi d}$$\Rightarrow \dfrac{\mu_0 il}{4\pi d \sqrt{d^2 + \dfrac{l^2}{4}}} = \dfrac{\mu_0 i}{\pi d}(\dfrac{1}{2} - \dfrac{1}{200})$$\Rightarrow \dfrac{l}{4 \sqrt{d^2 + \dfrac{l^2}{4}}} = \dfrac{99}{200}$ $\Rightarrow \dfrac{l^2}{d^2 + \dfrac{l^2}{4}} = (\dfrac{99\times 4}{200})^2 = \dfrac{156816}{40000}$ = 3.92$\\$$\Rightarrow l^2 = 3.92 d^2 + \dfrac{3.92}{4} l^2$$\\$$(\dfrac{1-3.92}{4})l^2 = 3.92 d^2 \Rightarrow 0.02 l^2 = 3.92 d^2 \Rightarrow 0.02 l^2 = 3.92 d^2 \Rightarrow \dfrac{d^2}{l^2} = \dfrac{0.02}{3.92} = \dfrac{d}{l} = \sqrt{\dfrac{0.02}{3.92}} = 0.07 17 17- Figure (35-E5) shows a square loop ABCD with edge- length a. The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the centre of the loop assuming uniform wires. ##### Solution : As resistances vary as r & 2r\\Hence Current along ABC = \frac{i}{3} & along ADC = \frac{2}{3i} Now, \\$$\vec{B}$ due to ADC = $2\Big[\dfrac{\mu_0 i\times 2\times 2\times \sqrt2}{4\pi 3a}\Big] = \dfrac{2\sqrt 2\mu_0 i}{3\pi a}$$\\$$\vec{B}$ due to ABC = $2\Big[\dfrac{\mu_0 i \times 2\times \sqrt 2}{4\pi 3a}\Big] = \dfrac{2\sqrt2 \mu_0 i}{6\pi a}$$\\Now \vec{B} = \dfrac{2\sqrt2 \mu_0i}{3\pi a} - \dfrac{2\sqrt2\mu_0 i}{6\pi a} = \dfrac{\sqrt2\mu_0 i}{3\pi a} 18 18- Figure (35-E6) shows a square loop of edge a made of a uniform wire. A current i enters the loop at the point A and leaves it at the point C. Find the magnetic field at the point P which is on the perpendicular bisector of AB at a distance a/4 from it. ##### Solution : A_0 = \sqrt{\dfrac{a^2}{16} + \dfrac{a^2}{4}} = \sqrt{\dfrac{5a^2}{16}} = \dfrac{a \sqrt 5}{4}$$\\$$D_0 = \sqrt{(\frac{3a}{4})^2 + (\frac{a}{2})^2} = \sqrt{\frac{9a^2}{16} + \frac{a^2}{4}} = \frac{a\sqrt {13}}{4}$$\\$Magnetic field due to AB$\\$$B_{AB} = \dfrac{\mu_0}{4\pi} \times \dfrac{i}{2(\dfrac{a}{4})}(sin(90-i) + sin(90-\alpha))$$\\$ = $\dfrac{\mu_0 \times 2i}{4\pi a}2 Cos \alpha = \dfrac{\mu_0 \times 2i}{4\pi a}\times 2\times \dfrac{(\frac{a}{2})}{a(\frac{\sqrt5}{4})} = \dfrac{2\mu_0 i}{\pi \sqrt5}$$\\Magnetic field due to DC\\$$B_{DC} = \dfrac{\mu_0}{4\pi} \times \dfrac{i}{2(\dfrac{3a}{4})}2sin(90 - B)$$\\= \dfrac{\mu_0i\times 4\times 2}{4\pi\times 3a} cos\beta = \dfrac{\mu_0 i}{\pi \times 3a}$$\times \dfrac{(\frac{a}{2})}{\frac{\sqrt{13a}}{4}} = \dfrac{2\mu_0i}{\pi a3\sqrt{13}}$$\\The magnetic field due to AD & BC are equal and appropriate hence cancle each other.\\Hence, net magnetic field is \dfrac{2\mu_0i}{\pi \sqrt5} - \dfrac{2\mu_0i}{\pi a3\sqrt{13}} = \dfrac{2\mu_0 i}{\pi a}\Big[\frac{1}{\sqrt5} - \frac{1}{3\sqrt{13}}\Big] 19 19- Consider the situation described in the previous problem. Suppose the current i enters the loop at the point A and leaves it at the point B. Find the magnetic field at the centre of the loop. ##### Solution : \vec{B} due t BC &\\$$\vec{B}$ due to AD at Pt ‘P’ are equal or Opposite$\\$Hence net $\vec{B}$ = 0 Similarly, due to AB & CD at P = 0$\\$ $\therefore$ The net $\vec{B}$ at the Centre of the square loop = zero.

20   20- The wire ABC shown in figure (35-E7) forms an equilateral triangle. Find the magnetic field B at the centre O of the triangle assuming the wire to be uniform.

##### Solution :

For AB$\ \ \ \$ B is along$\odot$$\ \ \ \ B = \dfrac{\mu_0 i}{4\pi r}(sin 60^{\circ} + sin 60^{\circ})$$\\$For AC$\ \ \ \$ B $\ \ \ \ $$\otimes$$ \ \ \ \$ $\dfrac{\mu_0 i}{4\pi r}(sin 60^{\circ} + sin 60^{\circ})$$\\ For BD \ \ \ \ B \ \ \ \$$\odot$$\ \ \ \ B =\dfrac{\mu_0 i}{4\pi r}(sin 60^{\circ})$$\\$For DC $\ \ \ \$ B $\ \ \ \$ $\otimes$ $\ \ \ \$ B = $\dfrac{\mu_0i}{4\pi r}(sin 60^{\circ})$$\\$$\therefore$ Net B = 0

21   21- A wire of length I is bent in the form of an equilateral triangle and carries an electric currrent i. (a) Find the magnetic field B at the centre, (b) If the wire is bent in the form of a square, what would be the value of B at the centre ?

##### Solution :

(a) $\Delta$ABC is Equilateral $\\$AB = BC = CA = $\frac{t}{3}$$\\current = i\\AO = \dfrac{\sqrt 3}{2}a = \dfrac{\sqrt 3\times l}{2\times 3} = \dfrac{l}{2\sqrt 3}$$\\$ $\phi_1 = \phi_2 = 60 ^{\circ}$$\\So, MO = \dfrac{l}{6\sqrt 3} as AM : MO = 2 : 1\\$$\vec{B}$ due to BC at <$\\$ = $\dfrac{\mu_0 i}{4\pi r}(sin \phi_1 + sin \phi_2) = \dfrac{\mu_0 i}{4\pi}\times i\times 6\sqrt 3\times \sqrt 3 = \dfrac{\mu_0i\times 9}{2\pi l}$$\\net \vec{B} = \dfrac{9\mu_0 i}{2\pi l}\times 3 = \dfrac{27\mu_0 i}{2\pi l}$$\\$

(b) $\vec{B}$ due to AD = $\dfrac{\mu_0 i\times 8}{4\pi l}\sqrt 2 = \dfrac{8\sqrt 2\mu_0 i}{4\pi l}$$\\Net \vec{B} = \dfrac{8\sqrt 2\mu_0 i}{4\pi l}\times 4 = \dfrac{8\sqrt 2\mu_0 i}{\pi l} 22 22- A long wire carrying a current i is bent to form a plane angle a. Find the magnetic field B at a point on the bisector of this angle situated at a distance x from the vertex. ##### Solution : Sin \dfrac{\alpha}{2} = \dfrac{r}{x}$$\\$$\Rightarrow r = x Sin(\frac{\alpha}{2})$$\\$Magnetic field B due to AR$\\$$\dfrac{\mu_0 i}{4\pi r}[Sin(180-(90-(\dfrac{\alpha}{2}))) + 1]$$\\$$\Rightarrow \dfrac{\mu_0 i[Sin(90-(\dfrac{\alpha}{2})) + 1]}{4\pi \times Sin(\dfrac{\alpha}{2})}$$\\$ = $\dfrac{\mu_0 i[Cos(\dfrac{\alpha}{2}) + 1]}{4\pi \times Sin(\dfrac{\alpha}{2})}$$\\= \dfrac{\mu_0 i2Cos^4(\frac{\alpha}{4})}{4\pi \times 2sin(\frac{\alpha}{4})Cos(\frac{\alpha}{4})} = \dfrac{\mu_0 i}{4\pi x}Cot (\frac{\alpha}{4})$$\\$The magnetic field due to both the wire.$\\$$\dfrac{2\mu_0 i}{4\pi x}Cot(\frac{\alpha}{4}) = \dfrac{\mu_0 i}{2\pi x}Cot(\frac{\alpha}{4}) 23 23- Find the magnetic field B at the centre of a rectangular loop of length I and width b, carrying a current i. ##### Solution : \vec{B} AB$$\\$$\dfrac{\mu_0\times i\times 2}{4\pi b}\times 2Sin\theta = \dfrac{\mu_0iSin\theta}{\pi b}$$\\$$\dfrac{\mu_0 il}{\pi b\sqrt{l^2 + b^2}} = \vec{B} DC \ \ \ \ \therefore$$Sin(l^2 + b)$ = $\dfrac{(\frac{l}{2})}{\sqrt{\frac{l^2}{4} + \frac{b^2}{4}}}$ = $\dfrac{l}{l^2 + b^2}$$\\$$\vec{B} BC$$\\$$\dfrac{\mu_0 i\times 2}{4\pi l}\times 2\times 2Sin\theta'$ = $\dfrac{\mu_0 iSin\theta'}{\pi l}$ $\ \ \ \$ $\therefore Sin \theta' = \dfrac{\frac{b}{2}}{\sqrt{\frac{l^2}{4} + \frac{b^2}{4}}} = \dfrac{b}{\sqrt{l^2 + b^2}}$$\\= \dfrac{\mu_0 ib}{\pi l\sqrt{l^2 + b^2}} = \vec{B}AD$$\\$Net $\vec{B}$ = $\dfrac{2\mu_0 il}{\pi b\sqrt{l^2 + b^2}} + \dfrac{2\mu_0 ib}{\pi l\sqrt{l^2 + b^2}} = \dfrac{2\mu_0 i(l^2 + b^2}{\pi lb\sqrt{l^2 + b^2}} = \dfrac{2\mu_0 i\sqrt{l^2 + b^2}}{\pi ib}$

24   24- A regular polygon of n sides is formed by bending a wire of total length 2nr which carries a current i. (a) Find the magnetic field B at the centre of the polygon, (b) By letting $n\rightarrow \infty$ deduce the expression for the magnetic field at the centre of a circular current.

$2\theta = \dfrac{2\pi}{n} \Rightarrow \theta = \dfrac{pi}{n} , l = \dfrac{2\pi r}{n}$$\\Tan \theta = \dfrac{l}{2x} \Rightarrow x = \dfrac{l}{2Tan\theta}$$\\$$\dfrac{l}{2} = \dfrac{\pi r}{n}$$\\$$B_{AB} = \dfrac{\mu_0 i}{4\pi x} (Sin\theta + Sin\theta) = \dfrac{\mu_0 i2Tan\theta \times 2Sin\theta}{4\pi l}$$\\$=$\dfrac{\mu_0 i2Tan(\frac{\pi}{n}) 2Sin(\frac{\pi}{n})n}{4\pi 2\pi r}$ = $\dfrac{\mu_0 inTan(\frac{\pi}{n})Sin(\frac{\pi}{n})}{2\pi^2 r}$$\\ For n sides, B_{net} = \dfrac{\mu_0 inTan(\frac{\pi}{n})Sin(\frac{\pi}{n})}{2\pi^2 r} 25 25- Each of the batteries shown in figure (35-E8) has an emf equal to 5 V. Show that the magnetic field B at the point P is zero for. any set of values of the resistances. ##### Solution : Net current in circuit = 0\\ Hence the magnetic field at point P = 0\\ [Owing to wheat stone bridge principle] 26 26- A straight, long wire carries a current of 20 A. Another wire carrying equal current is placed parallel to it. If the force acting on a length of 10 cm of the second wire is 2.0\times 10^{-5} N, what is the separation between them ? ##### Solution : Force acting on 10 cm of wire is 2\times 10^{-5} N\\$$\dfrac{dF}{dl} = \dfrac{\mu_0 i_1 i_2}{2\pi d}$$\\$$\Rightarrow \dfrac{2\times 10^{-5}}{10\times 10^{-2}} = \dfrac{\mu_0 \times 20\times 20}{2\pi d}$$\\$$\Rightarrow d = \dfrac{4\pi \times 10^{-7}\times 20\times 20\times 10\times 10^{-2}}{2\pi \times 2\times 10^{-5}} = 400\times 10^{-3} = 0.4 m = 40 cm$

27   27- Three coplanar parallel wires, each carrying a current of 10 A along the same direction, are placed with a separation 5.0 cm between the consecutive ones. Find the magnitude of the magnetic force per unit length acting on the wires.

i = 10 A$\\$ Magnetic force due to two parallel Current Carrying wires.$\\$$F = \dfrac{\mu_0 I_1 I_2}{2\pi r}$$\\$ so, $\vec{F}$ or 1 = $\vec{F}$ by 2 + $\vec{F}$ by 3$\\$$=\dfrac{\mu_0 \times 10\times 10}{2\pi \times 5\times 10^{-2}} + \dfrac{\mu_0 \times 10\times 10}{2\pi \times 10\times 10^{-2}}$$\\$= $\dfrac{4\pi \times 10^{-7}\times 10\times10}{2\pi \times 5\times 10^{-2}} + \dfrac{4\pi \times 10^{-7}\times 10\times 10}{2\pi \times 10\times 10^{-2}}$$\\= \dfrac{2\times 10^{-3}}{5} + \dfrac{10^{-3}}{5} = \dfrac{3\times 10^{-3}}{5} = 6\times 10^{-4} N towards \ \ \ middle wire 28 28- Two parallel wires separated by a distance of 10 cm carry currents of 10 A and 40 A along the same direction. Where should a third current be placed so that it experiences no magnetic force ? ##### Solution : \dfrac{\mu_0 10i}{2\pi x} = \dfrac{\mu_0 i40}{2\pi (10 - x)}$$\\$$\Rightarrow \dfrac{10}{x} = \dfrac{40}{10-x} \Rightarrow \dfrac{1}{x} = \dfrac{4}{10-x}$$\\$$\Rightarrow 10 – x = 4x \Rightarrow 5x = 10 \Rightarrow x = 2 cm\\ The third wire should be placed 2 cm from the 10 A wire and 8 cm from 40 A wire. 29 29- Figure (35-E9) shows a part of an electric circuit. The wires AB, CD and EF are long and have identical resistances. The separation between the neighbouring wires is 1.0 cm. The wires AE and BF have negligible resistance and the ammeter reads 30 A. Calculate the magnetic fcrce per unit length of AB and CD. ##### Solution : F_{AB} = F_{CD} + F_{EF}$$\\$ = $\dfrac{\mu_0 \times 10\times 10}{2\times 1\times 10^{-2}} + \dfrac{\mu_0 \times 10\times 10}{2\times 2\times 10^{-2}}$$\\ = 2\times 10^{-3} + 10^{-3} = 3\times 10^{-3} downward\\$$F_{CD} = F_{AB} = F_{EF}$$\\As F_{AB} & F_{EF} are equal and oppositely directed hence F = 0 30 30- A long, straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass density 1.0\times 10^{-4} kg/m is placed parallel to and directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight ? ##### Solution : \dfrac{\mu_0 i_1 i_2}{2\pi d} = mg(For a portion of wire of length 1m)\\$$\Rightarrow \dfrac{\mu_0\times 50\times i_2}{2\pi \times 5\times 10^{-3}} = 1\times 10^{-4}\times 9.8$$\\$$\Rightarrow \dfrac{4\pi \times 10^{-7} \times 5\times i_2}{2\pi \times 5\times 10^{-3}} = 9.8\times 10^{-4}$$\\$$\Rightarrow 2\times i_2\times 10^{-3} = 9.3\times 10^{-3}\times 10^{-1}$$\\$$\Rightarrow i_2 = \dfrac{9.8}{2}\times 10^{-1}$ = 0.49 A

31   31- A square loop PQRS carrying a current of 6.0 A is placed near a long wire carrying 10 A as shown in figure (35-E10). (a) Show that the magnetic force acting on the part PQ is equal and opposite to that on the part RS. (b) Find the magnetic force on the square loop.

$I_2$ = 6 A$\\$$I_1 = 10 A\\$$F_{PQ}$$\\'F' on dx = \dfrac{\mu_0 i_1 i_2}{2\pi x} dx = \dfrac{\mu_0 i_1 i_2}{2\pi} \dfrac{dx}{x} = \dfrac{\mu_0\times 30}{\pi}\dfrac{dx}{x}$$\\$$\vec{F}_{PQ} = \dfrac{\mu_0 \times 30}{x}\int_1\frac{dx}{x} = 30\times 4\times 10^{-7}\times [log x]_1^2$$\\$ $= 120\times 10^{-7} [log 3 - log 1]$$\\Similarly force of \vec{F}_{RS} = 120 × 10^{-7} [log 3 – log 1]$$\\$So, $\vec{F}_{PQ} = \vec{F}_{RS}$$\\$$\vec{F}_{PS} = \dfrac{\mu_0\times i_1 i_2}{2\pi \times 1\times 10^{-2}} - \dfrac{\mu_0\times i_1 i_2}{2\pi \times 2\times 10^{-2}}$$\\ = \dfrac{2\times 6\times 10\times 10^{-7}}{10^{-2}} - \dfrac{2\times 10^{-7}\times 6\times 6}{2\times 10^{-2}} = 8.4\times 10^{-4} N(Towards right)\\$$\vec{F}_{RQ} = \dfrac{\mu_0\times i_1 i_2}{2\pi \times 3\times 10^{-2}} - \dfrac{\mu_0 \times i_1 i_2}{2\pi\times 2\times 10^{-2}}$$\\$$ = \dfrac{4\pi \times 10^{-7}\times 6\times 10}{2\pi \times 3\times 10^{-2}} - \dfrac{4\pi \times 10^{-7}\times 6\times 6}{2\pi \times 2\times 10^{-2}} = 4\times 10^{-4} + 36\times 10^{-5} = 7.6\times 10^{-4} N$$\\Net force towards down\\ = (8.4 + 7.6) × 10^{-4} = 16 × 10^{-4} N 32 32- A circular loop of one turn carries a current of 5.00 A. If the magnetic field B at the centre is 0.200 mT, find the radius of the loop. ##### Solution : B = 0.2 mT, \ \ \ i = 5 A, \ \ \ n = 1, \ \ \ r = ?\\$$B = \dfrac{n\mu_0 i}{2r}$$\\$$\Rightarrow r = \dfrac{n\times \mu_0i}{2B} = \dfrac{1\times 4\pi \times 10^{-7}\times 5}{2\times 0.2\times 10^{-3}}$$\\ = 3.14 × 5 × 10 m = 15.7× 10 m = 15.7 × 10 cm = 1.57 cm\\ 33 33- A current-carrying circular coil of 100 turns and radius 5.0 cm produces a magnetic field of 6.0 \times 10^{-6} T at its centre. Find the value of the current. ##### Solution : B = \dfrac{n\mu_0 i}{2r}$$\\$n = 100,$\ \$r = 5cm = 0.05 m$\\$$\vec{B} = 6\times 10^{-5} T\\$$ i =\dfrac{2rB}{n\mu_0} = \dfrac{2\times 0.05\times 6\times 10^{-6}}{100\times 4\pi \times 10^{-7}} = \dfrac{3}{6.28}\times 10^{-1} = 0.0477 \approx 49 mA$

34   34- An electron makes $3\times 10^{5}$ revolutions per second in a circle of radius 0.5 angstrom. Find the magnetic field B at the centre of the circle.

$3\times 10^5$ revolutions in 1 sec.$\\$1 revolutions in $\dfrac{1}{3\times 10^5}$sec$\\$ $i = \dfrac{q}{t} = \dfrac{1.6\times 10^{-19}}{\big(\dfrac{1}{3\times 10^5}\big)}$ A$\\$$B = \dfrac{\mu_0 i}{2r} = \dfrac{4\pi \times 10^{-7}\times 1.6\times 10^{-19}\times 3\times 10^{5}}{2\times 0.5\times 10^{-10}}\dfrac{2\pi \times 1.6\times 3}{0.5}\times 10^{-11} = 6.028\times 10^{-10} = 6\times 10^{-10} 35 35- A conducting circular loop of radius a is connected to two long, straight wires. The straight wires carry a current i as shown in figure (35-E11). Find the magnetic field B at the centre of the loop. ##### Solution : I = \dfrac{i}{2} in each semicircle\\ABC = \vec{B} = \dfrac{1}{2} \times \dfrac{\mu_0(\frac{i}{2})}{2a} downwards\\ADC = \vec{B} = \dfrac{1}{2} \times \dfrac{\mu_0(\frac{i}{2})}{2a} upwards\\Net \vec{B} = 0 36 36- Two circular coils of radii 5.0 cm and 10 cm carry equal currents of 2.0 A. The coils have 50 and 100 turns respectively and are placed in such a way that their planes as well as the centres coincide. Find the magnitude of the magnetic field B at the common centre of the coils if the currents in the coils are (a) in the same sense (b) in the opposite sense. ##### Solution : r_1 = 5 cm \ \ \ \ r_2 = 10cm\\ n_1 = 50 \ \ \ \ n_2 = 100\\i = 2 A\\ (a) B = \dfrac{n_1\mu_0i}{2r_1} + \dfrac{n_2 \mu_0 i}{2r_2}$$\\$ = $\dfrac{50\times4\pi \times 10^{-7}\times 2}{2\times 5\times 10^{-2}} + \dfrac{100\times 4\pi\times 10^{-7}\times 2}{2\times 10\times 10^{-2}}$$\\$$ = 4\pi × 10^{-4} + 4\pi × 10^{-4} = 8\pi × 10^{-4}$$\\ (b) B = \dfrac{n_1 \mu_0 i}{2r_1} - \dfrac{n_2 \mu_0 i}{2r_2} = 0 37 37- If the outer coil of the previous problem is rotated through 90° about a diameter, what would be the magnitude of the magnetic field B at the centre ? ##### Solution : Outer Circle\\n = 100, \ \ \ r = 100m = 0.1m\\i = 2 A\\$$\vec{B} = \dfrac{n\mu_0i}{2a} = \dfrac{100\times 4\pi \times 10^{-7}\times 2}{2\times 0.1} = 4\pi \times 10^{-4}$ $\ \ \ \$horizontally towards West.$\\$Inner Circle r = 5 cm = 0.05 m, n = 50, i = 2 A$\\$$\vec{B} = \dfrac{n\mu_0 i}{2r} = \dfrac{4\pi \times 10^{-7}\times 2\times 50}{2 \times 0.05} = 4\pi \times 10^{-4} \ \ \ downwards\\Net B = \sqrt{(4\pi \times10^{-4})^2 + (4\pi \times 10^{-4})^2} = \sqrt{32\pi^2 \times 10^{-8}} = 17.7 \times 10^{-4} = 18\times 10^{-4} = 1.8\times 10^{-3} = 1.8mT Outer Circle\\n = 100, \ \ \ r = 100m = 0.1m\\i = 2 A\\$$\vec{B} = \dfrac{n\mu_0i}{2a} = \dfrac{100\times 4\pi \times 10^{-7}\times 2}{2\times 0.1} = 4\pi \times 10^{-4}$ $\ \ \ \$horizontally towards West.$\\$Inner Circle r = 5 cm = 0.05 m, n = 50, i = 2 A$\\$$\vec{B} = \dfrac{n\mu_0 i}{2r} = \dfrac{4\pi \times 10^{-7}\times 2\times 50}{2 \times 0.05} = 4\pi \times 10^{-4} \ \ \ downwards\\Net B = \sqrt{(4\pi \times10^{-4})^2 + (4\pi \times 10^{-4})^2} = \sqrt{32\pi^2 \times 10^{-8}} = 17.7 \times 10^{-4} = 18\times 10^{-4} = 1.8\times 10^{-3} = 1.8mT 38 38- A circular loop of radius 20 cm carries a current of 10 A. An electron crosses the plane of the loop with a speed of 2.0\times 10^{-8} \frac{m}{s}. The direction of motion makes an angle of 30° with the axis of the circle and passes through its centre. Find the magnitude of the magnetic force on the electron at the instant it crosses the plane. ##### Solution : r = 20 cm, \ \ i = 10 A, \ \ V = 2\times 10^6 \frac{m}{s}, \ \ \theta = 30°\\$$F = e(\vec{V}\times \vec{B} Sin\theta)$$\\ = 1.6\times 10^{-19}\times 2\times 10^6\times \dfrac{\mu_0i}{2r} Sin 30°$$\\$ = $\dfrac{1.6\times 10^{-19}\times 2\times 10^6\times 4\pi \times 10^{-7}\times 10}{2\times 2\times 20\times 10^{-2}} = 16\pi \times 10^{-19} N$

39   39- A circular loop of radius R carries a current I. Another circular loop of radius r ( << R) carries a current i and is placed at the centre of the larger loop. The planes of the two circles are at right angle to each other. Find the torque acting on the smaller loop.

##### Solution :

$\vec{B}$ Large loop = $\dfrac{\mu_0I}{2R}$$\\‘i’ due to larger loop on the smaller loop\\$$ = i(A × B) = i AB Sin 90° = i × \pi r^2 \times \frac{\mu_0 i}{2r}$

40   40- A circular loop of radius r carrying a current i is held at the centre of another circular loop of radius R(>> r) carrying a current I. The plane of the smaller loop makes an angle of 30° with that of the larger loop. If the smaller loop is held fixed in this position by applying a single force at a point on its periphery, what would be the minimum magnitude of this force ?

##### Solution :

The force acting on the smaller loop$\\$ $F = ilB Sin\theta$$\\ = \dfrac{i2\pi r\mu_0I1}{2R\times 2} = \dfrac{\mu_0iI\pi r}{2R}$$\\$

41   41- Find the magnetic field B due to a semicircular wire of radius 10.0 cm carrying a current of 5.0 A at its centre of curvature.

i = 5 Ampere, $\ \$ r = 10 cm = 0.1 m$\\$ As the semicircular wire forms half of a circular wire,$\\$So, $\vec{B} = \dfrac{1}{2}\dfrac{\mu_0 i}{2r} = \dfrac{1}{2} \times \dfrac{4\pi \times 10^{-7}\times 5}{2\times 0.1}$$\\ = 15.7 \times 10^{-6} T \approx 16 \times 10^{-6} T = 1.6 \times 10^{-5} T 42 42- A piece of wire carrying a current of 6.00 A is bent in the form of a circular Eire of radius 10.0 cm, and it subtends an angle of 120° at the centre. Find the magnetic field B due to this piece of wire at the centre. ##### Solution : B = \dfrac{\mu_0 i}{2R} \dfrac{\theta}{2r} = \dfrac{2\pi}{3\times 2\pi} \times \dfrac{\mu_0 i}{2R}$$\\$$= \dfrac{4\pi \times 10^{-7}\times 6}{6\times 10^{t10^{-2}}} = 4\pi \times 10^{-6}$$\\$ = $4\times 3.14\times 10^{-6} = 12.56\times 10^{-6} = 1.26\times 10^{-5} T$

43   43- A circular loop of radius r carries a current i. How should a long, straight wire carrying a current 4 i be placed in the plane of the circle so that the magnetic field at the centre becomes zero ?

##### Solution :

$\vec{B}$ due to loop$\dfrac{\mu_0 i}{2r}$$\\Let the straight current carrying wire be kept at a distance R from centre. Given I = 4i\\$$\vec{B}$ due to wire = $\dfrac{\mu_0I}{2\pi R} = \dfrac{\mu_0 \times 4i}{2\pi R}$$\\Now, the \vec{B} due to both will balance each other\\Hence \dfrac{\mu_0 i}{2r} = \dfrac{\mu_0 4i}{2\pi R} \Rightarrow R = \dfrac{4r}{\pi}$$\\$Hence the straight wire should be kept at a distance $\dfrac{4\pi}{r}$ from centre in such a way that the direction of current in it is opposite to that in the nearest part of circular wire. As a result the direction will $\vec{B}$ will be oppose.

44   44- A circular coil of 200 turns has a radius of 10 cm and carries a current of 2.0 A. (a) Find the magnitude of the magnetic field $\vec{B}$ at the centre of the coil, (b) At what distance from the centre along the axis of the coil will the field B drop to half its value at the centre ?

$n = 200, i = 2 A, r = 10 cm = 10 × 10^{-2}$ m$\\$(a) $B = \dfrac{n\mu_0 i}{2r} = \dfrac{200\times 4\pi \times 10^{-7}\times 2}{2\times 10\times 10^{-2}} = 2\times 4\pi \times 10^{-4}$$\\ = 2 × 4 × 3.14 × 10^{-4} = 25.12 × 10^{-4} T = 2.512 mT$$\\$$(b) B = \dfrac{n\mu_0 ia^2}{2(a^2 + d^2)^\frac{3}{2}} \Rightarrow \dfrac{n\mu_0 i}{4a} = \dfrac{n\mu_0 ia^2}{2(a^2 + d^2)^\frac{3}{2}}$$\\$$\Rightarrow \dfrac{1}{2a} = \dfrac{a^2}{2(a^2 + d^2)^\frac{3}{2}}$$ \ \ \$ $\Rightarrow (a^2 +d^2)^\frac{3}{2}2a^3$$\ \ \ \Rightarrow a^2 + d^2 = (2a^3)^\frac{2}{3}$$\\$$\Rightarrow a^2 + d^2 = (2^\frac{1}{3}a)^2$$ \ \$ $\Rightarrow (10^{-1})^2 + d^2 = 2^\frac{2}{3} (10^{-1})^2$$\\$$\Rightarrow 10^{-2} + d^2 = 2^\frac{2}{3} 10^{-2}$$\ \ \Rightarrow (10^{-2})(4^\frac{1}{3} - 1) = d^2$$\\$$\Rightarrow 10^{-2}(1.5874 - 1) = d^2 \ \ \Rightarrow d^2 = 10^{-2} \times 0.5874$$\\$$\Rightarrow d = \sqrt{10^{-2} \times 0.5874} = 10^{-1} \times 0.786 m = 7.66\times 10^{-2} = 7.66 cm 45 45- A circular loop of radius 4.0 cm is placed in a horizontal plane and carries an electric current of 5.0 A in the clockwise direction as seen from above. Find the magnetic field (a) at a point 3.0 cm above the centre of the loop (b) at a point 3.0 cm below the centre of the loop. ##### Solution : At O P the \vec{B} must be directed downwards\\ We Know B at the axial line at O & P\\ = \dfrac{\mu_0 ia^2}{2(a^2 + d^2)^\frac{3}{2}}$$ \ \ \$ a = 4 cm = 0.04 m$\\$$= \dfrac{4\pi \times 10^{-7} \times 5\times 0.0016}{2(0.0025)^\frac{3}{2}}$$ \ \ \ $$\\= 40\times 10^{-6} = 4 \times 10^{-5} T \ \ \ downwards in both the cases 46 46- A charge of 3.14\times 10^{-6}C is distributed uniformly over a circular ring of radius 20.0 cm. The ring rotates about its axis with an angular velocity of 60.0 \frac{rad}{s}. Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm from the centre. ##### Solution : q = 3.14\times 10^{-6}C, r = 20 cm = 0.2 m, w = 60\dfrac{rad}{sec}, \ \ \ i = \dfrac{q}{t} = \dfrac{3.14\times 10^{-6}\times 60}{2\pi \times 0.2} = 1.5\times 10^{-5}$$\\$$\dfrac{Electric field}{Magnetic field} = \dfrac{\dfrac{xQ}{4\pi \epsilon_0(x^2 + a^2)^\frac{3}{2}}}{\dfrac{\mu_0 ia^2}{2(a^2 + x^2)^\frac{3}{2}}}$$\\$$= \dfrac{9\times 10^9\times 0.05\times 3.14\times 10^{-6}\times 2}{4\pi \times 10^{-7}\times 15\times 10^{-5}\times (0.2)^2}$$\\$$= \dfrac{9\times 5\times 2\times 10^3}{4\times 13\times 4\times 10^{-12}} = \dfrac{3}{8} 47 47- A thin but long, hollow, cylindrical tube of radius r carries a current i along its length. Find the magnitude of the magnetic field at a distance r/2 from the surface (a) inside the tube (b) outside the tube. ##### Solution : (a) For inside the tube \ \ \vec{B} = 0\\ As, B inside the conducting tube = 0\\ (b) For \vec{B} outside the tube\\$$\dfrac{3r}{2}$$\\$$\vec{B} = \dfrac{\mu_0 i}{2\pi d} = \dfrac{\mu_0 i\times 2}{2\times \pi 3r} = \dfrac{\mu_0 i}{2\pi r}$

48   48- A long, cylindrical tube of inner and outer radii a and b carries a current i distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point (a) just inside the tube (b) just outside the tube.

##### Solution :

(a) At a point just inside the tube the current enclosed in the closed surface = 0. $\\$Thus B = $\dfrac{\mu_0 O}{A} = 0$$\\(b) Taking a cylindrical surface just out side the tube, from ampere’s law,\\ \mu_0 i = B\times 2\pi b$$ \ \ \$ $\Rightarrow B = \dfrac{\mu_0 i}{2\pi b}$

49   49- A long, cylindrical wire of radius b carries a current i distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point inside the wire at a distance a from the axis.

##### Solution :

i is uniformly distributed throughout.$\\$So, ‘i’ for the part of radius $a = \dfrac{i}{\pi b^2} \times \pi a^2 = \dfrac{ia^2}{b^2} = I$$\\Now according to Ampere’s circuital law\\$$\phi B\times dl = B\times 2\times \pi \times a = \mu_0 I$$\\$$\Rightarrow B = \mu_0 \dfrac{ia^2}{b^2}\times \dfrac{1}{2\pi a} = \dfrac{\mu_0 ia}{2\pi b^2}$

50   50- A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross-section. Find the magnetic field B at a point at a distance (a) 2 cm (b) 10 cm and (c) 20 cm away from the axis. Sketch a graph of B versus x for 0 < x < 20 cm.

##### Solution :

(a) r = 10 cm = $10\times 10^{-2} m$$\\ x = 2\times 10^{-2} m, \ \ \ i = 5 A\\i in the region of radius 2 cm\\= \dfrac{5}{\pi (10\times 10^{-2})^2} \times \pi(2\times 10^{-2})^2 = 0.2 A$$\\$$B\times \pi(2\times 10^{-2})^2 = \mu_0(0.2)$$\\$$\Rightarrow B = \dfrac{4\pi \times 10^{-7}\times 0.2}{\pi \times 4\times 10^{-4}} = \dfrac{0.2\times 10^{-7}}{10^{-4}} = 2\times 10^{-4}$$\\$(b) 10 cm radius $\\$$B\times \pi (10\times 10^{-2})^2 = \mu_0 \times 5$$\\$$\Rightarrow B = \dfrac{4\pi \times 10^{-7}\times 5}{\pi \times 10^{-2}} = 20 \times 10^{-5}$$\\$

(c) x = 20 cm$\\$$B\times \pi \times (20\times 10^{-2})^2 = \mu_0 \times 5$$\\$$\Rightarrow B = \dfrac{\mu_0 \times 5}{\pi \times (20\times 10^{-2})^2} = \dfrac{4\pi\times 10^{-7}\times 5}{\pi \times 400\times 10^{-4}} = 5\times 10^{-5} 51 51- Sometimes we show an idealised magnetic field which is uniform in a given region and falls to zero abruptly. One such field is represented in figure (35-E12). Using Ampere's law over the path PQRS, show that such a field is not possible. ##### Solution : We know, \int B\times dl = \mu_0 i. Theoritically B = 0 a t A\\If, a current is passed through the loop PQRS, then\\$$B = \dfrac{\mu_0i}{2(l + b)}$will exist in its vicinity.$\\$Now, As the $\vec{B}$ at A is zero. So there’ll be no interaction. However practically this is not true. As a current carrying loop, irrespective of its near about position is always affected by an existing magnetic field.

52   52- Two large metal sheets carry surface currents as shown in figure (35-E13). The current through a strip of width dl is Kdl where K is a constant. Find the magnetic field at the points P, Q and R.

##### Solution :

(a) At point P, i = 0, Thus B = 0$\\$

(b) At point R, i = 0, B = 0$\\$

(c) At point $\theta$, Applying ampere’s rule to the above rectangle$\\$$B \times 2I = \mu_0K_0 \int_0^1 dl$$\\$$\Rightarrow B \times 2I = \mu_0 kl \Rightarrow B = \dfrac{\mu_0 k}{2}$$\\$$B \times 2I = \mu_0 K_0 \int_0^1 dl$$\\$$\Rightarrow B\times 2I = \mu_0kI \Rightarrow B = \dfrac{\mu_0k}{2}Since the \vec{B} due to the 2 stripes are along the same direction, thus,\\$$B_{net} = \dfrac{\mu_0k}{2} + \dfrac{\mu_0k}{2} = \mu_0k$

53   53- Consider the situation of the previous problem. A particle having charge q and mass m is projected from the point Q in a direction going into the plane of the diagram. It is found to describe a circle of radius r between the two plates. Find the speed of the charged particle.

Charge = q,$\ \ \ \$mass = m$\\$We know radius described by a charged particle in a magnetic field B$\\$$r = \dfrac{mv}{qB}$$\\$Bit $B = \mu_0K$ [according to Ampere’s circuital law, where K is a constant]$\\$$r = \dfrac{mv}{q\mu_0k} \Rightarrow v = \dfrac{rq\mu_0k}{m} 54 54- The magnetic field B inside a long solenoid, carrying a current of 5.00 A, is 3.14\times10^{-2} T. Find the number of turns per unit length of the solenoid. ##### Solution : i = 25 A, B = 3.14 10^{-2} T, n =?\\$$B = \mu_0 ni$$\\$$\Rightarrow 3.14\times 10^{-2} = 4\times \pi \times 10^{-7} n\times 5$$\\$$\Rightarrow n = \dfrac{10^{-2}}{20\times 10^{-7}} = \dfrac{1}{2} \times 10^4 = 0.5\times 10^4 = 5000 \dfrac{turns}{m}$

55   55- A long solenoid is fabricated by closely winding a wire of radius 0.5 mm over a cylindrical nonmagnetic frame so that the successive turns nearly touch each other. What would be the magnetic field B at the centre of the solenoid if it carries a current of 5 A ?

r = 0.5 mm, i = 5 A, $B = \mu_0 ni$(for a solenoid)$\\$Width of each turn = 1 mm = $10^{-3}$ m$\\$No. of turns ‘n’ = $\dfrac{1}{10^{-3}} = 10^3$$\\So, B = 4\pi \times 10^{-7}\times 10^3\times 5 = 2\pi \times 10^{-3} T 56 56- A copper wire having resistance 0.01ohm in each metre is used to wind a 400-turn solenoid of radius 1.0 cm and length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of 1.0\times 10^{-2} T near the centre of the solenoid. ##### Solution : \dfrac{R}{I} = 0.01 \Omega in 1 m, \ \ r = 1.0 cm, \ \ Total turns = 400, \ \ l = 20cm,\\$$B = 1\times 10^{-2}$ T,$\ \$ $\dfrac{400}{20\times 10^{-2}} \dfrac{turns}{m}$$\\$$i = \dfrac{E}{R_0} = \dfrac{E}{R_0/I\times (2\pi r \times 400)} = \dfrac{E}{0.01\times 2\times \pi \times 0.01\times 400}$$\\$$B = \mu_0 ni$$\\$$\Rightarrow 10^2 = 4\pi \times 10^{-7}\times \dfrac{400}{20\times 10^{-2}}\times \dfrac{E}{400\times 2\pi \times 0.01\times 10^{-2}}$$\\$$\Rightarrow E = \dfrac{10^{-2}\times 20\times 10^{-2}\times 400\times 2\pi \times 10^{-2}\times 0.01}{4\pi \times 10^{-7}\times 400} = 1 V$

57   57- A tightly-wound solenoid of radius a and length I has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains n dx turns and may be approximated as a circular current i n dx. (a) Write the magnetic field at the centre of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the centre of the solenoid, (b) Verify that if l >> a, the field tends to $B = \mu_0 ni$ and if a >> l,the field tends to be $B = \dfrac{\mu_0 nil}{2a}$ Interpret these results.

##### Solution :

Current at ‘0’ due to the circular loop = dB = $\dfrac{\mu_0}{4\pi}\times \dfrac{a^2 indx}{\big[a^2 + (\dfrac{l}{2} - x)^2\big]^\frac{3}{2}}$$\\$$\therefore$ for the whole solenoid $B = \int_0^B dB$$\\ = \int_0^l \dfrac{a^2 nidx}{4\pi \big[a^2 + (\dfrac{l}{2} - x)^2\big]^\frac{3}{2}}$$\\$ $=\dfrac{\mu_0 ni}{4\pi} \int_0^l\dfrac{a^2 dx}{a^3\big[1 + (l - \dfrac{2x}{2a})^2 \big]^\frac{3}{2}}$$\\ = \dfrac{\mu_0 ni}{4\pi a} \int_0^l\dfrac{dx}{\big[1 + (l - \dfrac{2x}{2a})^2 \big]^\frac{3}{2}} = 1 + (l - \dfrac{2x}{2a})^2$$\\$

58   58- A tightly-wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of $1.00 \times 10^8 \frac{rev}{s}$. Find the number of turns per metre in the solenoid.

i = 2 a, f = $10 \frac{rev}{s}$, n = ?, $m_e = 9.1 \times 10^{-31} kg$,$\\$$q_e = 1.6 \times 10^{-19} c, B = \mu_0ni \Rightarrow n = \dfrac{B}{\mu_0i}$$\\$$f = \dfrac{qB}{2\pi m_e} \Rightarrow B = \dfrac{12\pi m_e}{q_e} \Rightarrow n = \dfrac{B}{\mu_0 i} = \dfrac{12\pi m_e}{q_e \mu_0 i} = \dfrac{10^8\times 9.1\times 10^{-31}}{1.6\times 10^{-19}\times 2\times 10^{-7}\times 2A} = 1421 \dfrac{turns}{m} 59 59- A tightly-wound, long solenoid has n turns per unit length, a radius r and carries a current i. A particle having charge q and mass m is projected from a point on the axis in a direction perpendicular to the axis. What can be the maximum speed for which the particle does not strike the solenoid ? ##### Solution : No. of turns per unit length = n, \ \ radius of circle = r/2, \ \ current in the solenoid = i,\\Charge of Particle = q, \ \ mass of particle = m \ \ \therefore B = \mu_0 ni$$\\$Again $\dfrac{mv^2}{r} = qVB \Rightarrow V = \dfrac{qBr}{m} = \dfrac{q\mu_0nir}{2m} = \dfrac{\mu_0 niqr}{2m}$

60   60- A tightly-wound, long solenoid is kept with its axis parallel to a large metal sheet carrying a surface current. The surface current through a width dl of the sheet is Kdl and the number of turns per unit length of the solenoid is n. The magnetic field near the centre of the solenoid is found to be zero, (a) Find the current in the solenoid, (b) If the solenoid is rotated to make its axis perpendicular to the metal sheet, what would be the magnitude of the magnetic field near its centre ?