Question

As shown in above figure, a uniform magnetic field B, direction of magnetic field is outward of the plane of the page. Calculate the potential difference between point a and b if metal rod is pulled upward with constant velocity.

• A. 0
• B. $\frac{1}{2}vBL$, with point a at the higher potential
• C. $\frac{1}{2}vBL$, with point b at the higher potential
• D. vBL, with point a at the higher potential
• E. vBL, with point b at the higher potential

Answer 1

The correct option is E vBL, with point b at the higher potential

When the rod is moved upward with velocity $v$ each electron $q$ inside the rod experiences a Lorentz magnetic force

$f_m=Bqv$ (given that magnetic field and velocity are perpendicular to each other)

it acts from point b to point a (by Fieming's left hand rule) ,due to this magnetic force electrons accumulate at the end a so end a becomes negative and end b positive.Hence an electric field is set up between the points b and a from b to a .The electric field so produced exerts electric force on electrons from a to b and opposes the magnetic Lorentz force. But due to accumulation of electrons at point a electric field increases hence electric force, after some time a stage comes when both forces become equal .If E is electric field at this stage then electric force on a electron

$f_e=qE$

in equilibrium both forces will be equal and opposite

$f_e=$$f_m$

$qE$=$Bqv$

$E=Bv$

if V be the potential difference and L is the length of rod , then

$E=V/L$

or we can say from last two equations ,

$V/L=vB$

$V=vBL$

As point is positive so it will be at higher potential.