Question

the force experienced by each pole of a magnet kept in the magnetizing field of strength 30N weber is 6×10^-3N the magnetic moment of length 10cm​

Answer 1

Explanation:

the force experienced by each pole of a magnet kept in the magnetizing field of strength 30N weber is 6×10^-3N the magnetic moment of length 10cm trending is 2

Answer 2

Given:

F = 30 N

B = 6×$10^{-3$  Wb

l = d = 10 cm = 0.1 m

To find:

M =?

Explanation:

The magnetic field strength of a straight current-carrying conductor is given by

$\displaystyle B = \frac{\mu_oI}{2\pi d}$     ...(1)

where

B = magnetic field strength

$\mu_0$ = permeability of free space

I = current

d = distance of the wire

Put given values in above eq(1)

$\displaystyle 6 \times10^{-3} = \frac{4\pi \times10^{-7} I}{2\pi \times 0.1}$

solving above eq, we get

I = 3 KA

The magnetic dipole moment is a vector quantity used to calculate the tendency of an object to interact with an external magnetic field.

Thus magnetic moment is given by

M = IA    ...(2)

where

M =  magnetic moment

I = current

A = area

Let's consider magnet is in cubic shape.

Area of magnet = 6 $l^2$

A = 6 ( 10 $\times 10^{-2})^2$

A = 6 × $10^{-2} \ m^2$

Put given values in above eq(2)

M = (3000) (6 $\times 10^{-2})$

M = 120 τ