First slide

Force on a current carrying wire placed in a magnetic field

Question

A 0.1 m long conductor carrying a current of 50A is perpendicular to a magnetic field of 1.25mT. The mechanical power to move the conductor with a speed of 1 m/s in a direction perpendicular to its length and perpendicular to its length and perpendicular to the magnetic field is,

Moderate

Solution

Magnetic force $F_{B} = ilB = 50 \times 0 .1 \times 1 .25 \times 10^{- 3} N = 6 .25 \times 10^{- 3} N$
$∴ Power F_{B} \times v = 6 .25 \times 10^{- 3} \times 1 J = 6 .25 mW$

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