Question

(Bar Magnet)
Suppose an isolated north pole is kept at the centre
of a circular loop carrying a electric current i. The
magnetic field due to the north pole at a point on the
periphery of the wire is B. The radius of the loop
is a. The force on the wire is
(1) Nearly 2naiB perpendicular to the plane of
the wire
(2) 2nail in the plane of the wire
(3) naiB along the axis of the wire
(4) Zero​

Answer 1

Answer:

Let us draw a circular loop in which current “i” is flowing in a direction as shown in the figure below.

The radius of this circular loop is given as “a”.

Then, a bar magnet is kept in the loop in such a way that it's north pole coincides with the centre of the loop. The magnetic field due to the north pole of the bar magnet will exert a force perpendicularly to the wire.

Now, let’s take a small section of this circular wire of length “dl”, assuming the original length of this wire to be “l”.

So, the current flowing across the section dl = i * dl

Therefore,  

The force acting on to section dl due to the magnet is,

dF = B * i * dl

on integrating, we get

∫dF = ∫Bidl ….. [taking limits from 0 to 2πa ∵ the circumference of the loop is 2πr and value of r is given as “a”]

⇒ F = B * i * ∫dl

⇒ F = B * i * 2πa

⇒ F = 2πaiB

Since the magnetic field due to the north pole of the bar magnet will be exerting force perpendicularly to the wire.

Thus, we can say option (1) is correct, which is the force on the wire is nearly “2πaiB” perpendicular to the plane of the wire.