Question

A short bar magnet is kept along the magnetic meridian with its north pole pointing north. A neutral point is located at a distance d from the centre. What is the ratio of magnetic inductions at a distance d/2 and 2d from the centre of the magnet on the equatorial line ?

• A. 9 : 64
• B. 8:1
• C. 9 : 56
• D. 64 : 9

Answer 1

Answer: (B)

8:1

mag. field at a distance x on equitorial line:
$B={\mu }_0/4\pi \times M/(x^2+l^2{)}^{3/2}$

Assuming $l<<x=>B=k\times M/x^3$

When north pole of magnet is aligned with geographical north, neutral points exists on the equatorial line.

At neutral point, $B=-B_H,x=d$

Net mag field = 0.

$B_d=\frac{kM}{d^3}=-B_H$

$B_{2d}=\frac{kM}{(2d{)}^3}=-B_H/8$

$B_{net,2d}=B_H+B_{2d}=7B_H/8$

Similarly,

$B_{d/2}=\frac{kM}{(d/2{)}^3}=-8B_H$

$B_{net,d/2}=B_H+B_{d/2}=-7B_H$

Finding the required ratio,

$\frac{B_{net,d/2}}{B_{net,2d}}=8$