First slide

Magnetic field due to current element

Question

A long, straight wire of radius a caries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wires at distance $\frac{a}{3}$ and 2a. respectively from the axis of the wire is :

Moderate

A

1/2
B

3/2
C

2/3
D

2

Solution

$\begin{array}{l} A t i n s i d e p o i n t P, B_{p} = \frac{μ_{0} i (\frac{a}{3})}{2 π a^{2}} = \frac{μ_{0} i}{6 π a} b e c a u s e i α π R^{2} a n d i (x) α π x^{2} \\ A t o u t s i d e p o i n t Q, B_{q} = \frac{μ_{0} i}{2 π (2 a)} \\ ∴ R a t i o = \frac{B_{p}}{B_{q}} = \frac{2}{3} \end{array}$

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