Question

A conducting rod AB of length $\ell =1m$ is moving at a velocity $v=4m/s$ making an angle ${30}^o$ with its length. A uniform magnetic field $B=2T$ existing in a direction perpendicular to the plane of motion. Then -

• A. $V_A-V_B=8V$
• B. $V_A-V_B=4V$
• C. $V_B-V_A=8V$
• D. $V_B-V_A=4V$

Answer 1

The correct option is D $V_B-V_A=4V$

Given,

Length $L=1m$

velocity, $v=4m/s$

Angle, $\theta ={30}^o$

Magnetic Field, $\stackrel{\rightharpoonup }{B}=2T$

Potential Difference,

$V=BLv\sin {30}^o=2\times 1\times 4\times \sin {30}^o=4V$

Potential difference is $4V$