First slide

Magnetism

Question

The magnetic potential at a point of a distance 40 cm from an isolated pole is $10^{- 5} Tm$ . Then Magnetic moment in $A m^{2}$ is

Moderate

A

12 $A m^{2}$
B

16 $A m^{2}$
C

10 $A m^{2}$
D

20 $A m^{2}$

Solution

Formula:
   $V = \frac{μ_{0}}{4 π} \frac{M \cos θ}{r^{2}}$
Given    $V = 10^{- 5} T m r = 40 c m \Rightarrow r=0.4m$
   $∴ 10^{- 5} = \frac{10^{- 7} M \cos 0^{o}}{16 \times 10^{- 2}} \Rightarrow M = 16 A m^{2}$

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