Question

A metallic bar PQ of length l, mass m and resistance R is placed on a frictionless metal rails of negligible resistance inclined at an angle θ above the horizontal. The uniform magnetic field B is in direction as shown in figure. The metallic bar is released from rest and slides down the rails and finally attains terminal velocity.

(1).The terminal velocity of the bar is : 

(a) \(\frac{mgR}{B^2l^2sin\theta}\)

(b) \(\frac{mgR\,sin\theta}{B^2l^2cos^2\theta}\)

(c) \(\frac{mgR\,cos\theta}{B^2l^2sin^2\theta}\)

(d) \(\frac{2mgR\,sin\theta}{B^2l^2cos^2\theta}\)

(2). The rate at which electrical energy is converted into heat energy in the bar after attaining terminal velocity is :

(a) \(\frac{B^2l^2v^2cos^2\theta}{R}\)

(b) \(\frac{B^2l^2v\,cos^2\theta}{R}\)

(c) \(\frac{B^2l^2v^2sin^2\theta}{R}\)

(d) \(\frac{B^2l^2vsin^2\theta}{R}\)

Answer 1

(1) Option : (b)

current (I) is induced in the bar due to changing magnetic field.

But will attain terminal velocity if the weight of the bar is completely balanced by the magnetic force acting on the bar.

i.e.,

(Bcosθ) Il = mg sinθ

But,

I = \(\frac{ɛ}{R}\)

= \(\frac{(B\,cos\,\theta)\,vl}{R}\)

∴ \(\frac{(B\,cos\,\theta)(B\,cos\,\theta)\,vl^2}{R}\) = mgsinθ

∴ Terminal velocity,

v = \(\frac{(mg\,sin\theta)R}{B^2l^2cos^2\theta}\)
 

(2) Option : (a)

H = I²R

= \((\frac{B\,cos\theta\,vl}{R})^2\) x R

= \(\frac{B^2l^2v^2cos^2\theta}{R}\)