First slide

Combined field of bar magnet and Earth; neutral point

Question

A short bar magnet of magnetic moment $1.28 \times 10^{- 2} {Am}^{2}$ is arranged in the magnetic meridian with its south pole pointing geographic north. If $B_{H} = 0.4$ G , the distance between the two null points is

Moderate

A

4cm
B

8cm
C

12cm
D

16cm

Solution

When a north pole of a bar magnet is facing south , null points are obtained on axial line
$G i v e n : B_{a} = B_{H}, B_{H} = 0.4 G = 4 \times 10^{- 5} T, M = 1.28 \times 10^{- 2} A m^{2}$
$W e k n o w, a t a x i s, B_{H} = \frac{μ_{0}}{4 π} \frac{2 M}{d^{3}}$
$\Rightarrow 4 \times 10^{- 5} = \frac{10^{- 7} \times 2 \times 1.28 \times 10^{- 2}}{d^{3}} \Rightarrow d^{3} = \frac{2 \times 1.28}{4} \times 10^{- 4}$
$\Rightarrow d^{3} = 64 \times 10^{- 6} \Rightarrow d = 4 \times 10^{- 2} m = 4 c m$
The distance between two null points
$2 d = 8 c m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App