Question

A short bar magnet of magnetic moment 10 A m2
is placed along x-axis at origin. The magnetic field
at (5 m, 0) is

(1) 0.8 * 10-8 T
(2) 1.6 x 10-8 T
(3) 10-8 T
(4) Zero​

Answer 1

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Answer 2

Answer:

The magnetic field at (5 m, 0) is 1.6×10⁻⁸T i.e.option(2).

Explanation:

The point 5 m is along the axis of the bar magnet.

And the magnetic field along the axis of the bar magnet is given as,

$B=\frac{\mu_o 2M}{4\pi r^3}$          (1)

Where,

B=magnetic field along the axis of the bar magnet

M=magnetic moment of the bar magnet

μ₀=permeability of free space=4π×10⁻⁷

r=distance of the point from the bar magnet

From the question we have,

M=10Am²

r=5m

By substituting the value of M and r in equation (1) we get;

$B=\frac{4\pi\times 10^-^7\times 2\times 10}{4\pi (5)^3}=0.16\times 10^-^7=1.6\times 10^-^8T$

Hence, the magnetic field at (5 m, 0) is 1.6×10⁻⁸T i.e.option(2).