First slide

Magnetic field due to a current carrying circular ring

Question

Two circular coils P and Q are made from similar wires but radius of Q is twice that of P. What should be the value of potential difference across them so that the magnetic induction at their centre may be the same ?

Easy

A

$V_{q} = 2 V_{p}$
B

$V_{q} = 3 V_{p}$
C

$V_{q} = 4 V_{p}$
D

$V_{q} = \frac{1}{4} V_{p}$

Solution

$B_{1} = \frac{μ_{0}}{4 π} (\frac{2 {πi}_{1}}{r_{1}}) and B_{2} = \frac{μ_{0}}{4 π} (\frac{2 {πi}_{2}}{r_{2}})$
Given that $r_{2} = 2 r_{1}, B_{1} = B_{2} ∴ i_{2} = 2 i_{1}$
$\frac{V_{q}}{V_{p}} = \frac{i_{2} \times r_{2}}{i_{1} \times r_{1}} = \frac{2 i_{1}}{i_{1}} \times \frac{2 r_{1}}{r_{1}} = \frac{4}{1}$

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