CBSE Class 12 · 2010 · Magnetic Effects of CurrentShort Answer

A square coil of side

10 cm

has 20 turns and carries a current of

12 A

. The coil is suspended vertically and the normal to its plane makes an angle of

3 0^{\circ}

with the direction of a uniform horizontal magnetic field. If the torque experienced by the coil equals

0.96 N\cdotm

, find the magnitude of the magnetic field.

Answer: $B = 0.8 T$ .

Solution

Area

A = (0.10)^{2} = 0.01 m^{2}

. Torque on a coil:

τ = N I A B sin θ

0.96 = 20 \times 12 \times B \times 0.01 \times sin 3 0^{\circ}

0.96 = 20 \times 12 \times 0.01 \times \frac{1}{2} \times B = 1.2 B

B = \frac{0.96}{1.2} = 0.8 T

CBSE Class 12 Board · 2010 · Subjective