# NCERT : Moving Charges & Magnetism

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 , I1 = 8A , I2 = 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 = μo 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 m2

τ = n B I A sin⁡α

τ =20 × 0.80 × 12 × (0.1)2 × sin⁡30°

= 0.96 Nm

Q:10. Two moving coil metre M1 and M2 have the following particulars:

R1=10Ω ;N1 =30 ;A1 =3.6×10-3 m2 ;B1 = 0.25 T;

R2=14Ω,N2=42,A2 = 1.8×10-3 m2 ; B2 = 0.50 T

(The spring constants are identical for the two metres).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1

Sol.
For meter M1, R1 = 10 Ω ; N1 = 30 ; A1 = 3.6 × 10-3 m2 ; B1 = 0.25 T ; k1 = k

For meter M2, R2 = 14 Ω, N2 = 42 ; A2 = 1.8 ×10-3 m2 ; B2 = 0.50 T ; k2 = 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 $\displaystyle V_s = \frac{N B A }{k R}$

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

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

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

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

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