Question

A very thin 19.0 cm copper bar is aligned horizontally along the east-west direction. If it moves horizontally from south to north at velocity = 11.0 m/s in a vertically upward magnetic field and B = 1.18 T , what potential difference is induced across its ends ? which end (east or west) is at a higher potential ? a) East b) West

Answer 1

Answer:

2.47 V,East

Explanation:

We are given that

l=19 cm=$19\times 10^{-2} m$

$1 cm=10^{-2} m$

$v=11 m/s$

B=1.18 T

We have to find the potential difference induced across its ends.

$E=Bvl$

Using the formula

$E=1.18\times 11\times 19\times 10^{-2}$

$E=2.47 V$

Hence, the potential difference induces across its ends=2.47 V

The positive charge  will move towards east direction and the negative charge will move towards west direction because the direction of force will be east.Therefore, the potential at east end will be high.