A conducting rod `PQ` of length `l = 2m` is moving at a speed of `2m s^(-1)` making an angle of `30^(@)` with its length. A uniform magnetic field `B = 2T` exists in a direction perpendicular to the plane of motin. Then

Options

A. (a) `V_(P) - V_(Q) = 8V`
B. (b) `V_(P) - V_(Q) = 4 V`
C. ( c) `V_(Q) - V_(P) = 8 V`
D. ( d) `V_(Q) - V_(P) = 4 V`

Answer

B

Explanation

(b) Emf induced across the rod `AB` is
`e = vec(B).(vec(l) xx vec(v))`
`= Blv sin theta`
`= 2 xx 2 xx 2 xx sin30`
`e = 4 V`
Free electrons of the rod shift towards roght due to force ltbgt `q(vec(v) xx vec(B))`
Thus, end `P` is at higher potential
or `V-(P) - V_(Q) = 4V`. Thus, choice (b) is correct.