Moving Charges and Magnetism

Physics

NCERT

1   A circular coil of wire consisting of $100$ turns, each of radius $8.0 cm$ carries a current of $0.40 A$. What is the magnitude of the magnetic field B at the centre of the coil?

Solution :

Number of turns on the circular coil, $n = 100$$\\$ Radius of each turn,$ r = 8.0 cm =0.08 m$$\\$ Current flowing in the coil, $I = 0.4 A$$\\$ Magnitude of the magnetic field at the centre of the coil is given by the relation,$\\$ $|B|=\dfrac{\mu_0}{4\pi}\dfrac{2\pi nl}{r}$$\\$ $\mu_0=$Permeability of free space$\\$ $=4\pi * 10^{-7}TmA^{-1}\\ |B|=\dfrac{4\pi *10^{-7}}{4 \pi} * \dfrac{2 \pi * 100 * 0.4}{0.08}\\ =3.14*10^{-4}T$$\\$ Hence, the magnitude of the magnetic field is $3.14*10^{-4} T .$

2   A long straight wire carries a current of $35 A.$ What is the magnitude of the field $B$ at a point $20 cm$ from the wire?

Solution :

Current in the wire,$ I =35 A$$\\$ Distance of a point from the wire, $r = 20 cm = 0.2 m$$\\$ Magnitude of the magnetic field at this point is given as:$\\$ $B=\dfrac{\mu_0}{4\pi}\dfrac{2l}{r}$$\\$ Where,$\\$ $\mu_0=$Permeability of free space$=4\pi*10^{-7}TmA^{-1}$$\\$ $B=\dfrac{4\pi*10^{-7}*2*35}{4\pi *0.2}\\ =3.5*10^{-5}T$$\\$ Hence, the magnitude of the magnetic field at a point $20 cm$ from is $3.5 *10^{-5} T .$

3   A long straight wire in the horizontal plane carries a current of $50 A$ in north to south direction. Give the magnitude and direction of $B$ at a point $2.5 m$ east of the wire.

Solution :

Current in the wire, $I = 50 A$$\\$ A point is $2.5 m$ away from the East of the wire.$\\$ $\therefore $ Magnitude of the distance of the point from the wire,$ r =2.5 m$ .$\\$ Magnitude of the magnetic field at that point is given by the relation,$\\$ $B=\dfrac{\mu_0 2l}{4\pi r}$$\\$ Where,$\\$ $\mu_0=$Permeability of free space$=4\pi*10^{-7}TmA^{-1}$$\\$ $B=\dfrac{4\pi*10^{-7}*2*50}{4\pi*2.5}\\ =4*10^{-6}T$$\\$ The point is located normal to the wire length at a distance of $2.5 m$. The direction of the current in the wire is outward. Hence, according to the Maxwell’s right and thumb rule, the direction of the magnetic field at the given point is vertically upward.

4   A horizontal overhead power line carries a current of $90 A$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5 m $ below the line?

Solution :

Current in the power line, $I = 90 A$$\\$ Point is located below the power line at distance, $r = 1.5 m$$\\$ Hence, magnetic field at that point is given by the relation,$\\$ $B=\dfrac{\mu_02l}{4 \pi r}$$\\$ Where,$\\$ $\mu_0=$Permeability of free space$=4\pi*10^{-4}TmA^{-1}$$\\$ $B=\dfrac{4 \pi * 10^{-7}*2 * 90}{4 \pi *1.5}\\ =1.2*10^{-5}T$$\\$ The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.

5   What is the magnitude of magnetic force per unit length on a wire carrying a current of $8 A$ and making an angle of $30^o$ with the direction of a uniform magnetic field of $0.15 T$?

Solution :

Current in the wire,$ I = 8 A$$\\$ Magnitude of the uniform magnetic field, $B = 0.15 T$$\\$ Angle between the wire and magnetic field, $\theta=30^o$$\\$ Magnetic force per unit length on the wire is given as:$\\$ $f=BI \sin \theta \\ =0.15* 8 * 1* \sin 30^o\\ =0.6 N m^{-1}$$\\$ Hence, the magnetic force per unit length on the wire is $0.6 N m^{-1} $.

6   A $3.0 cm$ wire carrying a current of $10 A$ is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27 T$. What is the magnetic force on the wire?

Solution :

Length of the wire,$ l = 3 cm = 0.03 m$$\\$ Current flowing in the wire,$I = 10A$$\\$ Magnetic field, $B = 0.27 T$$\\$ Angle between the current and magnetic field, $\theta 90 ^o $( Because magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis) Magnetic force exerted on the wire is given as: $F = BIl \sin \theta \\ =0.27 *10 *0.03\sin90 ^o\\ =8.1 * 10 ^{- 2} N$$\\$ Hence, the magnetic force on the wire is $8.1 * 10^{- 2} N $. The direction of the force can be obtained from Fleming’s left hand rule.

7   Two long and parallel straight wires A and B carrying currents of $ 8.0 A $ and $ 5.0 A $ in the same direction are separated by a distance of $ 4.0 cm.$ Estimate the force on a $ 10 cm $ section of wire A.

Solution :

Current flowing in wire $A, I_ A =8.0A$$\\$ Current flowing in wire $B, I_ B = 5.0A$$\\$ Distance between the two wires,$ r = 4.0 cm = 0.04 m$$\\$ Length of a section of wire $A, l = 10 cm = 0.1 m$$\\$ Force exerted on length l due to the magnetic field is given as:$\\$ $B=\dfrac{\mu _0 2I_A I_B l}{4 \pi r}$$\\$ Where,$\\$ $\mu 0 =$ Permeability of free space $= 4 \pi * 10^{-7 } T mA ^{-1}$$\\$ $B=\dfrac{4 \pi *10^{-7} * 2 * 8 * 5 * 0.1}{4 \pi * 0.04}\\ =2* 10^{-5}N$$\\$ The magnitude of force is $ 2 * 10^{- 5} N$ . This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

8   A closely wound solenoid $80 cm$ long has $5$ layers of windings of $400$ turns each. The diameter of the solenoid is $1.8 cm $. If the current carried is $8.0 A$, estimate the magnitude of B inside the solenoid near its centre.

Solution :

Length of the solenoid,$ l = 80 cm = 0.8 m$$\\$ There are five layers of windings of $400$ turns each on the solenoid. $\therefore $ Total number of turns on the solenoid, $N = 5 * 400 = 2000$$\\$ Diameter of the solenoid, $D = 1.8 cm = 0.018 m$$\\$ Current carried by the solenoid,$ I = 8.0A$$\\$ Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,$\\$ $ B=\dfrac{\mu _0 NI}{l}$$\\$ Where, $\\$ $\mu _0 = $ Permeability of free space $=4 \pi * 10^{- 7} T mA ^{-1}$$\\$ $B=\dfrac{4 \pi * 10^{-7}*2000 * 8}{0.8}\\ =8 \pi * 10 ^{-3}=2.512*10^{-2}T$$\\$ Hence, the magnitude of the magnetic field inside the solenoid near its centre is $2.512*10^{-2}T$

9   A square coil of side $10 cm $ consists of $20 $ turns and carries a current of $12 A.$ The coil is suspended vertically and the normal to the plane of the coil makes an angle of $30^o$ with the direction of a uniform horizontal magnetic field of magnitude $0.80 T$. What is the magnitude of torque experienced by the coil?

Solution :

Length of a side of the square coil,$ l = 10 cm = 0.1 m$$\\$ Current flowing in the coil, $I = 12A$$\\$ Number of turns on the coil, $n = 20$$\\$ Angle made by the plane of the coil with magnetic field, $\theta 30 ^o$$\\$ Strength of magnetic field, $B = 0.80T$$\\$ Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,$\\$ $\tau = n BIA \sin \theta \\ \implies l * l = 0.1 * 0.1 = 0.01 m ^2\\ \therefore \tau =20 *0.8 * 12 * 0.01 * \sin30 ^o\\ =0.96 N m$$\\$ Hence, the magnitude of the torque experienced by the coil is $0.96 N m.$

10   Two moving coil meters, $M _1$ and $M _2$ have the following particulars:$\\$ $R_ 1 = 10 \Omega , N _1 = 30,$$\\$ $A _1 = 3.6 * 10 ^{- 3} m^ 2 , B_ 1 = 0.25T$$\\$ $R _2 = 14 \Omega , N_ 2 = 42$$\\$ $A_ 2 = 1.8 * 10^{-3 }m ^2 , B_ 2 = 0.50T$$\\$ (The spring constants are identical for the meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of $M_ 2$ and $M _1 $.

Solution :

For moving coil meter $M _1 :$$\\$ Resistance, $R_ 1 = 10 \Omega $$\\$ Number of turns, $N _1 = 30$$\\$ Area of cross-section, $A_ 1 = 3.6 * 10^{- 3} m ^2$$\\$ Magnetic field strength, $B_ 1 = 0.25T$$\\$ Spring constant $K_ 1 = K$$\\$ For moving coil meter $M _2 :$$\\$ Resistance, $R _2 =14 \Omega $$\\$ Number of turns, $N_ 2 = 42$$\\$ Area of cross-section, $A _2 = 1.8 * 10^{- 3} m^ 2$$\\$ Magnetic field strength, $B_ 2 = 0.50T$$\\$ Spring constant $K_ 2= K$$\\$

(a) Current sensitivity of $M_ 1$ is given as:$\\$ $I_{s1}=\dfrac{N_1 B_1 A_1}{K_1}$$\\$ And, current sensitivity of $M_ 2$ is given as:$\\$ $I_{s2}=\dfrac{N_2 B_2 A_2}{K_2}$$\\$ $\therefore $ Ratio =$\dfrac{I_{s2}}{I_{s1}}\\ =\dfrac{N_2 B_2 A_2 K_1}{K_2 N_1B_1A_1}$$\\$ $\therefore $ Ratio $ =\dfrac{42 * 0.5 * 1.8 * 10^{-3}*10 * K}{k*14*30*0.25*3.6*10^{-1 } }=1.4$$\\$ Hence, the ratio of current sensitivity of $M_ 2$ of $M _1$ is $1.4.$$\\$ (b) Voltage sensitivity for $M_ 2$ is given is: $\\$ $=V_{s2}=\dfrac{N_2 B_2 A_2}{K_2 R_2}$$\\$ And, voltage sensitivity for $M_ 1$ is given as:$\\$ $V_{s1}=\dfrac{N_1 B_1 A_1}{K_1}$$\\$ $\therefore $ Ratio $ \dfrac{V_{s2}}{V_{s1}}=\dfrac{N_2B_2A_2K_1R_1}{K_2R_2N_1B_1A_1}$$\\$ $=\dfrac{42*0.5*1.8*10^{-3}*10*K}{K*14*30*0.25 * 3.6 *10^{-3}}=1$$\\$ Hence, the ratio of voltage sensitivity of $M_ 2 $ to $M_ 1$ is $ 1.$