Question 269: Magnetism and Matter

Question

A circular coil of radius 2 cm and 125 turns carries a current of 1 A . The coil is placed in a uniform magnetic field of magnitude 0.4 T . The axis of the coil makes an angle of $30^{\circ}$ with the direction of the magnetic field. The torque acting on the coil is $\alpha 	imes 10^{-4} \mathrm{~N} . \mathrm{m}$. The value of $\alpha$ is $\_\_\_\_$ .
$$ (\pi=3.14) $$

Accepted Answer

For a current-carrying coil placed in a uniform magnetic field, the torque is
$$ 	au = N I A B \sin 	heta $$
where:
$N = 125$
$I = 1 \, 	ext{A}$
$B = 0.4 \, 	ext{T}$
radius $r = 2 \, 	ext{cm} = 0.02 \, 	ext{m}$
$	heta = 30^\circ$
Here, $	heta$ is the angle between the axis of the coil and the magnetic field.
Now area of the circular coil:
$$ A = \pi r^2 = 3.14 	imes (0.02)^2 $$
$$ A = 3.14 	imes 0.0004 = 0.001256 \, 	ext{m}^2 $$
Now substitute in the torque formula:
$$ 	au = 125 	imes 1 	imes 0.001256 	imes 0.4 	imes \sin 30^\circ $$
Since,
$$ \sin 30^\circ = \frac{1}{2} $$
So,
$$ 	au = 125 	imes 0.001256 	imes 0.4 	imes 0.5 $$
$$ 	au = 125 	imes 0.001256 	imes 0.2 $$
$$ 	au = 125 	imes 0.0002512 $$
$$ 	au = 0.0314 \, 	ext{N m} $$
Now write it in the form $\alpha 	imes 10^{-4} \, 	ext{N m}$:
$$ 0.0314 = 314 	imes 10^{-4} $$
Therefore,
$$ \alpha = 314 $$
So, the answer is:
$$ \boxed{314} $$

Magnetism and Matter

Integer / Numeric Answer Type