A conducting bar pulled with a constant speed v on a smooth conducting rail. The region has a steady magnetic field of induction B as shown in the figure. If the speed of the bar is doubled then the rate of heat dissipation will be
Text Solution
Verified by ExpertsThe correct answer is:
C
The induced emf between A and B = ε ε = B l λ v
⇒ ⇒ The induced current = i = 
⇒ ⇒ i = 
The electrical power = P = i 2 R = 
Since v is doubled, the electrical power, becomes four times. Since heat dissipation per second is proportional to electrical power, it becomes four fold.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The current in a coil changes from 4 ampere to zero in 0.1 s. If the average e.m.f. induced is 100 …
If in a coil rate of change of area is second and current become 1 amp from 2 amp in . If magnitu…
An e.m.f. of 12 volt is produced in a coil when the current in it changes at the rate of 45 amp/min…
An average induced e.m.f. of 1V appears in a coil when the current in it is changed from 10A in one…
If a change in current of 0.01 A in one coil produces a change in magnetic flux in the other coil,…
A solenoid of length l meter has self-inductance L henry. If number of turns are doubled, its self-…