Q.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 ?

Sol. Given , N = 100 , r = 0.08m , I= 0.4A

$ \displaystyle B =\frac{\mu_0}{4\pi}\frac{2\pi N I}{r} $

B = 10^{−7}×2×3.14×0.4×100/0.08

Q2. 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 ?

Sol. By formula ,

$ \displaystyle B =\frac{\mu_0}{4\pi}\frac{2I}{r} $

B = 10^{−7}×2×35/0.20

= 3.5 × 10^{−5} T

Q3. 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.

Sol. By formula ,

$ \displaystyle B =\frac{\mu_0}{4\pi}\frac{2I}{r} $

B = 10^{−7}×2×50/2.5

B = 4 × 10^{−6} T

Q4. 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 ?

Sol. By formula ,

$ \displaystyle B =\frac{\mu_0}{4\pi}\frac{2I}{r} $

B = 10^{−7}×2×90/1.5

B = 1.2×10^{−5} T

Q5. What is the magnitude of magnetic force per unit length on a wire

carrying a current of 8 A and making an angle of 30º with the

direction of a uniform magnetic field of 0.15 T ?

Sol. By formula , F = IlBsinθ

F = 8×1×0.15×Sin30º

F = 0.6 N/m

Q6. 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 ?

Sol. By formula , F = IlBsinθ

F = 10×0.03×0.27×Sin90º

F = 8.1×10^{−2} N

Q7. 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.

Sol. By using formula of ,

Force per unit length $ \displaystyle F = \frac{\mu_0}{4\pi}\frac{2I_1 I_2}{r} $

Given , I_{1} = 8A , I_{2} = 5A , r = 0.04 m

Force acting on length l is = F x l

Q8. 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 .

Sol. l = 0.8m , N= 5×400 = 2000 , I = 8A

no. of turns per unit length , n = 2000/0.8 = 2500

Magnetic field inside the solenoid is B = μ_{o} nI

After putting values ,

B = 4π × 10^{−7} × 2500 × 8 T

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

Sol.

Here, l = 10 cm = 0.10 m ; n=20 ; I=12 A ; α =30° ;B=0.08 T ; τ = ?

Area, A = l×l = 0.10×0.10=(0.1)^{2} m^{2}

τ = n B I A sinα

τ =20 × 0.80 × 12 × (0.1)^{2} × sin30°

= 0.96 Nm

Q:10. Two moving coil metre M_{1} and M_{2} have the following particulars:

R_{1}=10Ω ;N_{1} =30 ;A_{1} =3.6×10^{-3} m^{2} ;B_{1} = 0.25 T;

R_{2}=14Ω,N_{2}=42,A_{2} = 1.8×10^{-3} m^{2} ; B_{2} = 0.50 T

(The spring constants are identical for the two metres).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_{2} and M_{1}

Sol.

For meter M_{1}, R_{1} = 10 Ω ; N_{1} = 30 ; A_{1} = 3.6 × 10^{-3} m^{2} ; B_{1} = 0.25 T ; k_{1} = k

For meter M_{2}, R_{2} = 14 Ω, N_{2} = 42 ; A_{2} = 1.8 ×10^{-3} m^{2} ; B_{2} = 0.50 T ; k_{2} = k.

As , Current sensitivity $ \displaystyle I_s = \frac{N B A }{k}$

(a) $ \displaystyle \frac{I_{s_2}}{I_{s_1}} = \frac{N_2 B_2 A_2 }{N_1 B_1 A_1}$

$ \displaystyle \frac{I_{s_2}}{I_{s_1}} = \frac{42 \times 0.50 \times 1.8 \times 10^{-3} }{30 \times 0.25 \times 3.6 \times 10^{-3}}$

$\displaystyle \frac{I_{s_2}}{I_{s_1}} = 1.4 $

(b) Voltage sensitivity $ \displaystyle V_s = \frac{N B A }{k R}$

$ \displaystyle \frac{V_{s_2}}{V_{s_1}} = \frac{N_2 B_2 A_2 /k R_2}{N_1 B_1 A_1/k R_1}$

$ \displaystyle \frac{V_{s_2}}{V_{s_1}} = \frac{N_2 B_2 A_2 R_1 }{N_1 B_1 A_1 R_2}$

$ \displaystyle \frac{V_{s_2}}{V_{s_1}} = \frac{42 \times 0.50 \times 1.8 \times 10^{-3}\times 10 }{30 \times 0.25 \times 3.6 \times 10^{-3} \times 14}$

$ \displaystyle \frac{V_{s_2}}{V_{s_1}} = 1 $