An iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now its magnetic moment will be

Question

An iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now its magnetic moment will be (A) M (B) (2 M/π) (C) (M/π) (D) M π

Tardigrade Admin · Accepted Answer

On bending a rod its pole strength remains unchanged whereas its magnetic moment changes. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fd15b01c20b2d82c503a96942932f821-.png" /> New magnetic moment M'=m(2 R)=m ((2 L/π)) ≡ (2 M/π)

Q. An iron rod of length $L$ and magnetic moment $M$ is bent in the form of a semicircle. Now its magnetic moment will be

$M$

$\frac{2 M}{π}$

$\frac{M}{π}$

$M π$

On bending a rod its pole strength remains unchanged whereas its magnetic moment changes.

New magnetic moment $M^{'} = m (2 R) = m$
$(\frac{2 L}{π}) \equiv \frac{2 M}{π}$

Q. An iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now its magnetic moment will be

M

π2M​

πM​

Mπ

On bending a rod its pole strength remains unchanged whereas its magnetic moment changes. New magnetic moment M′=m(2R)=m (π2L​)≡π2M​

Q. An iron rod of length $L$ and magnetic moment $M$ is bent in the form of a semicircle. Now its magnetic moment will be

$M$

$\frac{2 M}{π}$

$\frac{M}{π}$

$M π$

On bending a rod its pole strength remains unchanged whereas its magnetic moment changes.

New magnetic moment $M^{'} = m (2 R) = m$
$(\frac{2 L}{π}) \equiv \frac{2 M}{π}$