Question

Magnetic induction due to a short bar magnet on its axial line is inversely proportional to cube of distance of the point.

• A. True
• B. False

Answer 1

The correct option is A True

Here Ab is short bar magnet with,

Length=2l

Pole strength=$\pm m$

Magnetic dipole moment $M=m\times 2l$

Let P be at 'd' distance from O thus,

Magnetic field at P due to (+m) is

$B_1=\frac{{\mu }_0}{4\pi }\frac{m}{(d-l{)}^2}$ (1)

Similarly due to (+m)

$B_2\frac{{\mu }_0}{4\pi }\frac{-m}{(d+l{)}^2}$ (2)

Now total magnetic field due to +m and (m) at P is $B=B_1+B_2$

$⟹B=\frac{{\mu }_{0}m}{4\pi }(\frac{1}{{(d-l)}^2}-\frac{1}{{(d+l)}^2})$

$⟹B=\frac{{\mu }_{0}m4dl}{4\pi {(d^2-l^2)}^2}$

Neglecting length of magnet, being short

$B=\frac{{\mu }_0(m\times 2l)2d}{4\pi {(d^2-l^2)}^2}$

$B=\frac{{\mu }_0{M}_2}{4\pi d^3}$

Thus$B\propto \frac{1}{d^2}$