A metal bar of mass m slides frictionlessly on two parallel conducting rails a d
ID: 1414189 • Letter: A
Question
A metal bar of mass m slides frictionlessly on two parallel conducting rails a distance d apart. A resistor R is connected across the rails and a uniform magnetic field B, pointing into the page, fills the entire region.
(a) If the bar moves to the right at speed v, what is the current in the circuit? In what direction does it flow?
(b) What is the magnetic force on the bar? In what direction?
(c) If the bar starts out with speed v0 at time t=0, and is left to slide, what will happen at later time? What is the total energy delivered to the resistor?
Explanation / Answer
we know that lenz law is consistent with law of conservation of energy we can explore this aspect with the motion of a conductor in a magnetic field
suppose a conductor is moved with a constant velocity v on parallel sides of a U shaped conductor in a magnetic field B let R be the resistance of a closed loop
the emf induced in the rod => e=Bdv
(a) ans
the current in the circuit i=Bdv/R
as current flows in the conductor from bottom to top
(b) ans
the magnetic force acts on the conductor Fm=[B^2d^2v]/R
the direction of the force is opposite to the velocity of a conductor
(c) ans
the total energy delivered to resistor is E=Fm*d=B^2d^3V/R