Question

A small coil of N turns has an effective area A and carries a current I. It is suspended in a horizontal magnetic field $\vec B$ such that its plane is perpendicular to $\vec B$ . The work done in rotating it by ${180^0}$ about the vertical axis is
(A) $NAIB$
(B) $2NAIB$
(C) $2\pi NAIB$
(D) $4\pi NAIB$

Answer 1

Hint:First start with what is work done in general and how it works then try to substitute the general condition with the case of coil suspended in a horizontal magnetic field. Use all the information provided in the question in this specific condition of the coil suspended in the magnetic field and get the required answer.

Complete answer:
First start with all the information given in the question:
Number of turns in the coil is N.
Effective area of the coil is A.
Current carrying in the coil is I.

Angle between the plane of the coil and the magnetic field is ${0^0}$ .
So, ${\theta _1} = {0^0}$

The coil is rotated by ${180^0}$ about the vertical axis.
So, ${\theta _2} = {180^0}$

Now we know the work done in case of coil suspended in the magnetic field and it is rotated to certain angle is given by;
$W = MB(\cos {\theta _1} - \cos {\theta _2})$
Where, $M = NIA$
Then, $W = NIA(B\cos {0^0} - \cos {180^0})$
$W = 2NAIB$

Hence the correct answer is Option(B).

Note: Be careful about the angle of plane of coil made with the magnetic field here it was perpendicular to each other but it is not same all the time and the answer will change accordingly. Also the angle of rotation will also change in every other question and hence the work done will get change too.