Question

An isosceles right angled current carrying loop PQR is placed in
uniform magnetic field 1 pointing along PR. If the magnetic force acting
on the arm PQ in F, then the magnetic force which acts on the small
(A
F
car
(D)
-F​

Answer 1

Given :

Isosceles right angled triangle PQR,

Magnetic field along PR.

Force on arm PQ = F

To find :

Force along QR.

Solution :

Ampere's circuital law :

A law which states that the integral of magnetic field density along an imaginary closed path is equal to the product of permeability of the medium and current enclosed by the path

$\int B\cdot dl = \mu_0 \times I \:$

In case of a current carrying loop, the total current goes outside and inside the loop is same, so

The net force on this loop is equal to 0.

Force along PR is also zero as magnetic field is along PR.

So

It is given that

Force along arm PQ = F

Force along arm PR = 0

so, Net force in current carrying loop is:

$F_{net} = F_{PQ} \: + F_{QR} \: + F_{PR} \:$

We now know that

net force in current carrying loop is zero,

And putting values of all other forces,

we get

$0= F \: + F_{QR} \: + 0\:$

So,

Net force along arm QR :

$\bf F_{QR} = - F \:$

So,

Net force along arm QR is equal to -F.