question_answer

At a place the value of horizontal component of the earth's magnetic field H is \[3\times {{10}^{-5}}Wb/{{m}^{2}}.\] A metallic rod AB of length 2m placed in east-west direction, having the end A towards east, falls vertically downward with a constant velocity of 50 m/s. Which end of the rod becomes positively charged and what is the value of induced potential difference between the two ends?

A)  \[EndA,3\times {{10}^{-3}}mV\]              

B)  End A, 3 mV

C)  \[End\text{ }B,3\times {{10}^{-3}}mV\]               

D)  End B, 3 mV

Correct Answer: D

Solution :

Induced potential difference in the rod. \[e=Bvl\]                 or            \[e=Hvl\] Given:  \[H=3\times {{10}^{-5}}Wb/{{m}^{2}},\] \[v=50m/s,\]\[l=2m\]                 \[\therefore \]  \[e=3\times {{10}^{-5}}\times 50\times 2\]                 \[=3\times {{10}^{-3}}V\]                 \[=3mV\] Now, by Right hand palm rule. If H is towards N and rod is moving downward, then the current will move form B to A. Hence, end B will become positive.