First slide

Magnetic field due to a current carrying circular ring

Question

In figure, AB is a non-conducting rod. Equal charges of magnitude q are fixed at various points on the rod as shown. The rod is rotated uniformly about an axis passing through O and perpendicular to its length such that linear speed of the end A or B of the rod is 3 mis. Magnetic field at O is

Difficult

A

$\frac{11 μ_{0} q}{12 π}$
B

$\frac{3 μ_{0} q}{7 π}$
C

$\frac{μ_{0} q}{2 π}$
D

$\frac{6 μ_{0} q}{13 π}$

Solution

Since the rod is a rigid body, every point has same angular velocity $ω = 1 rad / s .$
For points at a distance of 1 m, v₁ = 1 m/s
For points at a distance of 2 m, v₂ = 2 m/s and for points at a distance of 3m , $v_{3}$ =3 $\frac{m}{s}$
If a charge rotating in a circle constitutes a current I,
$B = \frac{μ_{0} i}{2 r}, where i \frac{q}{t} = \frac{qv}{2 πr}$
Since field due to rotation of all charges are in same direction
$B = \sum \frac{μ_{0} qv}{4 {πr}^{2}} = \frac{μ_{0} q}{4 π} (1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \frac{1}{3})$
$\Rightarrow B = \frac{11 μ_{0} q}{12 π}$

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