# Permanent Magnets

## Concept Of Physics

### H C Verma

1   A long bar magnet has a pole strength of 10 A-m. Find the magnetic field at a point on the axis of the magnet at a distance of 5 cm from the north pole of the magnet.

##### Solution :

$m = 10 A-m, \qquad d = 5 cm = 0.05 m$ $\\$ $B = \frac {\mu_0 m}{4 \pi r^{2}} = \frac{10^{-7} \times 10}{(5 \times 10^{-2})^2} = \frac{10^{-2}}{25} = 4 \times 10^{-4} \ Tesla$

2   Two long bar magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2.0 cm from the south pole of the second. If both the magnets have a pole strength of 10 A-m, find the force exerted by one magnet on the other.

##### Solution :

m1 =m2 = 10 A-m $\\$ r = 2 cm = 0.02 m $\\$ we know $\\$ Force exerted by tow magnetic poles on each other =$\frac{\mu_0}{4 \pi} \frac{m_1m_2}{r^2} = \frac{4 \pi \times 10^{-7} \times 10^2}{4 \pi \times 4 \times 10^{-4}} = 2.5 \times 10^{-2} \ N$

3   A uniform magnetic field of $0.20 \times 10^{-3} T$ exists in the space. Find the change in the magnetic scalar potential as one moves through 50 cm along the field.

##### Solution :

$B = - \frac{dv}{dl} \Rightarrow dv = -B dl = - 0.2 \times 10^{-3} \times 0.5 = -0.1 \times 10^{-3} T-m$ $\\$ Since the sigh is –ve therefore potential decreases.

4   Figure (36-E1) shows some of the equipotential surfaces of the magnetic scalar potential. Find the magnetic field B at a point in the region.

##### Solution :

Here $dx = 10 \ sin 30° \ cm = 5 cm$ $\\$ $\frac{dV}{dx} = B =\frac{0.1 \times 10^{-4} T-m}{5 \times 10^{-2} m}$ $\\$ Since B is perpendicular to equipotential surface. $\\$ Here it is at angle 120° with (+ve) x-axis and B = 2 × 10–4 T

5   The magnetic field at a point, 10 cm away from a magnetic dipole, is found to be 2.0 x 10 T. Find the magnetic moment of the dipole if the point is (a) in end-on position of the dipole and (b) in broadside-on position of the dipole.

##### Solution :

$B = 2 × 10^{–4} T$ $\\$ $d = 10 cm = 0.1 m$ $\\$ (a) if the point at end-on postion. $\\$ $B = \frac{ \mu_0}{4 \pi} \frac{2M}{d^{3}} \Rightarrow 2 \times 10^{-4} = \frac{10^{-7} \times 2M}{(10^{-1})^3}$ $\\$ $\Rightarrow \frac{2 \times 10^{-4} \times 10^{-3}}{10^{-7} \times 2} = M \Rightarrow M= 1 \ Am^2$ $\\$ b) If the point is at broad-on position $\\$ $\frac{ \mu_0}{4 \pi} \frac{M}{d^{3}} \Rightarrow 2 \times 10^{-4} = \frac{10^{-7} \times M}{(10^{-1})^3} \Rightarrow M = 2 \ A m^2$

6   Show that the magnetic field at a point due to a magnetic dipole is perpendicular to the magnetic axis if the line joining the point with the centre of the dipole makes an angle of $tan^{-1}(\sqrt2)$ with the magnetic axis

##### Solution :

Given : $\\$ $\theta = tan^{-1} \sqrt2 \Rightarrow tan \theta = \sqrt 2 \Rightarrow 2 = tan^2 \theta$ $\\$ $tan \theta = 2 cot \theta \Rightarrow \frac{tan \theta}{2} = cot \ \theta$ $\\$ We know $\frac{tan \theta}{2} = tan \ \alpha$ $\\$ Comparing we get, $tan \alpha = cot \theta$ $\\$ or,$tan \alpha = tan(90 – \theta) \qquad or \alpha = 90 – \theta \qquad or \theta + \alpha = 90$ $\\$ Hence magnetic field due to the dipole is $\bot$r to the magnetic axis.

7   A bar magnet has a length of 8 cm. The magnetic field at a point, at a distance 3 cm from the centre in the broadside-on position is found to be $4 \times 10^{-6} \ T$. Find the pole strength of the magnet.

##### Solution :

Magnetic field at the broad side on position : $\\$ $B = \frac{\mu_0}{4 \pi}\frac{M}{(d^{2} + l^{2})^3/2} \qquad 2l =8cm \qquad d=3cm$ $\\$ $\Rightarrow 4 \times 10^{-6} = \frac{10^{-7} \times m \times 8 \times 10^{-2}}{(9 \times 10^{-4} + 16 \times 10^{-4})^3/2} \Rightarrow 4 \times 10^{--6} = \frac{10^{-9} \times m \times 8}{(10^{-4})^{3/2} + (25)^{3/2}}$ $\\$ $\Rightarrow m = \frac{4 \times 10^{-6} \times 125 \times 10^{-8}}{8 \times 10^{-9}} = 62.5 \times 10^{-5} A-m$

8   A magnetic dipole of magnetic moment 1.44 A-m is placed horizontally with the north pole pointing towards north. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 $\mu$T.

##### Solution :

We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on position.$\\$ Again B in this case = $\frac{\mu_0 m}{4 \pi d^3}$ $\\$ $theefore \frac{\mu_0 m}{4 \pi d^3} = \overrightarrow{B_h}$ due to earth $\\$ $\Rightarrow \frac{10^{-7} \times 1.44}{d^3} = 18 \mu T$ $\\$ $\Rightarrow \frac{10^{-7} \times 1.44}{d^3} = 18 \times 10^{-6}$ $\\$ $\Rightarrow d^3 =8 \times 10^{-3}$ $\\$ $\Rightarrow d = 2 × 10^{–1} m = 20 cm$ $\\$ In the plane bisecting the dipole.

9   A magnetic dipole of magnetic moment $0.72 A-m^2$ is placed horizontally with the north pole pointing towards south. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18$\mu$T.

##### Solution :

When the magnet is such that its North faces the geographic south of earth. The neutral point lies along the axial line of the magnet. $\\$ $\frac{\mu_0}{4 \pi} \frac{2M}{d^3} = 18 \times 10^{-6} \Rightarrow \frac{10^{-7} \times 2 \times 0.72}{d^3} = 18 \times 10^{-6} \Rightarrow d^3 = \frac{2 \times 0.7 \times 10^{-7}}{18 \times 10^{-6}}$ $\\$ $\Rightarrow d = (\frac{8 \times 10^{-9}}{10^{-6}})^{1/3} = 2 \times 10^{-1} m =20 \ cm$

10   A magnetic dipole of magnetic moment $0.72\sqrt 2 \ A-m ^2$ is placed horizontally with the north pole pointing towards east. Find the position of the neutral point if the horizontal component of the earth's magnetic field is $18 \mu T$.

##### Solution :

Magnetic moment = $0.72 \sqrt 2 A-m^2 = M$ $\\$ $B = \frac{\mu_0}{4 \pi} \frac{M}{d^3} \qquad B_h =18 \mu T$ $\\$ $\Rightarrow \frac{4 \pi \times 10^{-7} \times 0.72 \sqrt 2}{4 \pi \times d^3} = 18 \times 10^{-6}$ $\\$ $\Rightarrow d^3 = \frac{0.72 \times 1.414 \times 10^{-7}}{18 \times 10^{-6}} = 0.005656$ $\\$ $\Rightarrow d \approx 20cm$

11   The magnetic moment of the assumed dipole at the earth's centre is $8.0 \times 10 22 A-m^2$. Calculate the magnetic field B at the geomagnetic poles of the earth. Radius of the earth is 6400 km.

##### Solution :

The geomagnetic pole is at the end on position of the earth.$\\$ $B =\frac{\mu_0}{4 \pi} \frac{2 \pi}{d3} = \frac{10^{-7} \times 2 \times 8 \times 10^{-22}}{(6400 \times 10^{3})^{3}} \approx 60 \times 10^{-6}T =60 \mu T$

12   If the earth's magnetic field has a magnitude $3.4 \times 10^5 T$ at the magnetic equator of the earth, what would be its value at the earth's geomagnetic poles ?

##### Solution :

$\overrightarrow B= 3.4 × 10^{–5} T$ $\\$ Given $\frac{\mu_0}{4 \pi} \frac{M}{R^3} = 3.4 \times 10^{-5}$ $\\$ $\Rightarrow M=\frac{3.4 \times 10^{-5} \times R^3 \times 4 \pi}{4 \pi \times 10^{-7}} = 3.4 \times 10^2R^3$ $\\$ $\overrightarrow B$ at Poles = $\frac{\mu_0}{4 \pi} \frac{2M}{R^3} == 6.8 \times 10^{-5}T$

13   The magnetic field due to the earth has a horizontal component of $26 \mu T$ at a place where the dip is 60°. Find the vertical component and the magnitude of the field.

##### Solution :

$\delta(dip) = 60^0$ $\\$ $BH = B cos 60°$ $\\$ $\Rightarrow B = 52 × 10^{–6} = 52 \mu T$ $\\$ $B_V =B sin \delta =52 \times 10^{-6} \frac {\sqrt{3}}{2} = 44.98 \mu T \approx 45 \mu T$

14   A magnetic needle is free to rotate in a vertical plane which makes an angle of. 60° with the magnetic meridian. It' the needle stays in a direction making an angle of $\tan^{-1}(2/\sqrt{3})$ with the horizontal, what would be the dip at that place ?

##### Solution :

If $\delta_1$ and $\delta_2$ be the apparent dips shown by the dip circle in the 2$\bot$r positions, the true dip $\delta$ is given by $\\$ $cot^2 \delta = cot^2 \delta_1 + cot^2 \delta_2$ $\\$ $\Rightarrow cot^2 \delta =cot^2 45° + Cot^2 53°$ $\\$ $\Rightarrow cot^2 \delta = 1.56 \Rightarrow \delta =38.6 \approx 39°$

15   The needle of a dip circle shows an apparent dip of 45° in a particular position and 53° when the circle is rotated through 90°. Find the true dip.

##### Solution :

We know $\qquad B_H = \frac{\mu_0 in}{2r}$ $\\$ Give :$B_H = 3.6 \times 10^{-5}T \qquad \theta = 45°$ $\\$ $i = 10 mA = 10^{-2}A \qquad tan \theta = 1$ $\\$ $n = ? \qquad r =10 \ cm = o.1m$ $\\$ $n = \frac{B_H tan \theta \times 2r}{\mu_0 i } = \frac{3.6 \times 10^{-5} \times 2 \times 1\times 10^{-1}}{4 \pi \times 10^{-7} \ times 10^{-2}} = 0.5732 \times 10^{3} \approx 573 \ turns$

16   A tangent galvanometer shows a deflection of 45° when 10 mA of current is passed through it. If the horizontal component of the earth's magnetic field is $B_H= 3.6 \times10^{-5} T$ and radius of the coil is 10 cm, find the number of turns in the coil.

##### Solution :

$n = 50 \qquad A=2cm \times 2cm = 2 \times 2 \times 10^{-4}m^{2}$ $\\$ $i =20 \times 10^{-3}A \qquad B=0.5T$ $\\$ $\tau = ni(\overrightarrow{A} \times \overrightarrow{B}) = niAB \ sin 90° = 50 \times 20 \times 10^{-3} \times 4 \times 10^{-4} \times 0.5 = 2 \times 10^{-4}N-M$

17   A moving-coil galvanometer has a 50-turn coil of size 2 cm x 2 cm. It is suspended between the magnetic poles producing a magnetic field of 0.5 T. Find the torque on the coil due to the magnetic field when a current of 20 mA passes through it.

##### Solution :

$Given \theta = 37° \qquad d=10 cm =0.1m$ $\\$ We know $\\$ $\frac{M}{B_H} = \frac{4 \pi}{\mu_0} \frac{(d^2 - l^2)^{2}}{2d} tan \theta = \frac{4 \pi}{\mu_0} \times \frac{d^{4}}{2d} tan \theta$ [As the magnet is short] $\\$ $=\frac{4 \pi}{4 \pi \times 10^{-7}} \times \frac{(0.1)^{3}}{2} \times \tan 37° = 0.5 \times 0.75 \times 1 \times 10^{-3} \times 10^{-7} = 0.375 \times 10^{4}=3.75 \times 10^{4} = 3.75 \times 10^{4} = 3.75 \times 10^3 A-m^2 T^{-1}$

18   A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth's horizontal magnetic field.

##### Solution :

$\frac{M}{B_H}$(found in the previous problem) = $3.75 ×10^{3} A-m^{2} T^{–1}$ $\\$ $\theta = 37° \qquad d = ?$ $\\$ $\frac{M}{B_H} = \frac{4 \pi}{\mu_0}(d^2 + l^2)^{3/2}tan \theta$ $\\$ l << d $\qquad$ neglecting l w.r.t.d $\\$ $\Rightarrow \frac{M}{B_H} = \frac{4 \pi}{\mu_0}d^3Tan \theta \Rightarrow 3.75 \times 10^{3} = \frac{1}{10^{-7}} \times d^{3} \times 0.75$ $\\$ $\Rightarrow d^3 = \frac{3.75 \times 10^{3} \times 10^{-7}}{0.75} = 5 \times 10^{-4}$ $\\$ $\Rightarrow d = 0.079 m = 7.9cm$

19   The magnetometer of the previous problem is used with the same magnet in Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle ?

##### Solution :

Given $\frac{M}{B_H} = 40 A-M^2/T$ $\\$ Since the magnet is short $'\ell'$ can be neglected So, $\frac{M}{B_H} =\frac{4 \pi}{\mu_0} \times \frac{d^3}{2} =40$ $\\$ $\Rightarrow d^3 = \frac{40 \times 4 \pi \times 10^{-7} \times 2}{4 \pi} = 8\times 10^{-6}$ $\\$ $\Rightarrow d= 2 \times 10^{-2} m =2 cm$ $\\$ with the northpole pointing towards south.

20   A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having M/B„ = 40 A-m 2/T be placed so that the needle can stay in any position ?

##### Solution :

$\\$ According to oscillation magnetometer, $\\$ $T = 2 \pi \sqrt{\frac{I}{MB_H}}$ $\\$ $\Rightarrow \frac{\pi}{10} = 2 \pi \sqrt{\frac{1.2 \times 10^{-4}}{M \times 30 \times 10^{-6}}}$ $\\$ $\Rightarrow (\frac{1}{20})^2 = \frac{1.2 \times 10^{-4}}{M \times 30 \times 10^{-6}}$ $\\$ $\Rightarrow M=\frac{1.2 \times 10^{-4} \times 400}{30 \times 10^{-6}} = 16 \times 10^{2} A-m^2 = 1600 A-m^2$

21   A bar magnet takes it/10 second to complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is $1.2 \times 10^{-4} kg-m^2$ and the earth's horizontal magnetic field is 30 pT. Find the magnetic moment of the magnet.

##### Solution :

$\\$ We know $\upsilon = \frac{1}{2 \pi}\sqrt{\frac{mB_H}{I}}$ $\\$ For like poles tied together $\\$ M = M1 – M2 $\\$ For unlike poles M'= M1 + M2 $\\$ $\frac{\upsilon_1}{\upsilon_2} =\sqrt{\frac{M_1 - M_2}{M_1 + M_2}} \Rightarrow (\frac{10}{2})^2 = \frac{M_1- M_2}{M_1 + M_2} \Rightarrow25 = \frac{M_1 - M_2}{M_1 + M_2}$ $\\$ $\Rightarrow \frac{26}{24} = \frac{2M_1}{2M_2}\Rightarrow \frac{M_1}{M_2} =\frac{13}{12}$

22   The combination of two bar magnets makes 10 oscillations per second in an oscillation magnetometer when like poles are tied together and 2 oscillations per second when unlike poles are tied together. Find the ratio of the magnetic moments of the magnets. Neglect any induced magnetism.

##### Solution :

$\\$ $B_H = 24 \times 10^{-6}T \qquad T_1 = 0.1'$ $\\$ $B = B_H - B_{wire} =2.4 \times 10^{-6} -\frac{\mu_0}{2 \pi} \frac{i}{r} =24 \times 10^{-6} - \frac{2 \times 10^{-7} \times 18}{0.2} = (24-10) \times 10^{-6} =14 \times 10^{-6}$ $\\$ $T =2 \pi \sqrt{\frac{I}{MB_H}} \qquad \frac{T_1}{T_2} =\sqrt{\frac{B}{B_H}}$ $\\$ $\frac{0.1}{T_2} = \sqrt{\frac{14 \times 10^{-6}}{24 \times 10^{-6}}} = (\frac{0.1}{t_2})^2 = \frac{14}{24} \Rightarrow T_2^2 =\frac{0.01 \times 14}{24} \Rightarrow T_2 =0.076$

23   A short magnet oscillates in an oscillation magnetometer with a time period of 0.10 s where the earth's horizontal magnetic field is 24 pT. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.

##### Solution :

$\\$ $T = 2 \pi \sqrt{\frac{I}{MB_H}} \qquad Here I' =2I$ $\\$ $T_1 = \frac{1}{40}min \qquad T2=?$ $\\$ $\frac{T_1}{T_2} = \sqrt{\frac{I}{I'}}$ $\\$ $\frac{1}{40T_2} =\sqrt{\frac{1}{2}} \Rightarrow \frac{1}{ ^{2}} =\frac{1}{2} \Rightarrow T_2^2 =\frac{1}{800} \Rightarrow T_2 =0.03536$ $\\$ For 1 oscillation Time taken = 0.03536 min. $\\$ For 40 Oscillation Time = $4 × 0.03536 = 1.414 = \sqrt{2}$min

24   A bar magnet makes 40 oscillations per minute in an oscillation magnetometer. An identical magnet is demagnetized completely and is placed over the magnet in the magnetometer. Find the time taken for 40 oscillations by this combination. Neglect any induced magnetism.

##### Solution :

$\\$ $\gamma_1 = 40 \ oscillations/minute$ $\\$ $B_H =25 \mu T$ $\\$ m of second magnet = 1.6 $A-m^2$ $\\$ $d = 20 cm = 0.2 m$ $\\$ (a) For north facing north $\\$ $\gamma = \frac{1}{2 \pi}\sqrt{\frac{MB_H}{I}} \qquad \frac{1}{2 \pi}\sqrt{\frac{M(B_H - B}{I}}$ $\\$ $B = \frac{\mu_0}{4 \pi}\frac{m}{d^3} = \frac{10^{-7}\times 1.6}{8 \times 10^{-3}} = 20 \mu T$ $\\$ $\frac{\gamma_1}{\gamma_2} = \sqrt{\frac{B}{B_H - B}} \Rightarrow \frac{40}{\gamma_2} = \sqrt{\frac{25}{5}} \Rightarrow \gamma_2 = \frac{40}{\sqrt{5}} = 17.88 \approx 18 \ osci/min$ $\\$b) For north pole facing south $\\$ $\gamma_1 = \frac{1}{2 \pi}\sqrt{\frac{MB_H}{I}} \qquad \gamma_2 = \frac{1}{2 \pi}\sqrt{\frac{M(B_H - B)}{I}}$ $\\$ $\frac{\gamma_1}{\gamma_2} = \sqrt{\frac{B}{B_H - B}}\Rightarrow \frac{40}{\gamma_2} = \sqrt{\frac{25}{5}} \Rightarrow \gamma_2 = \frac{40}{\sqrt{(\frac{25}{45})}} = 53.66 \approx 54 \ osci/min$

25   A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is $25 \mu T$. Another short magnet of magnetic moment $1.6 A-m^2$ is placed 20 cm east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north pole (a) towards, north and (b) towards south.