A 20 cm long conducting rod is st into pure translation with a uniform velocity of 10 cm s–1 perpendicular to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion.

(a) Find the average magnetic force on the free electrons of the rod.


(b) For what electric field inside the rod, the electric force on a free electron will balance the magnetic force? How is this electric field created?


(c) Find the motional emf between the ends of the rod.



Given:


Length of rod l=20cm=0.2m


Velocity of rod v


Magnetic field intensity


(a) we know that,


A charge q moving with velocity v inside a magnetic field B experiences a force F given by


…. (i)


Since velocity of rod is perpendicular to magnetic field the above eqn. reduces to



To find the force on a free electron


Putting the values of q, v and B we get,



Therefore average force experienced by a free electron is


(b) we know that,


force experienced by a charge particle having charge q in presence of an electric field E is given by



To balance this force with the magnetic force, equating above eqn. with eqn.(i),




Putting the values of v and B we get,



This electric field is created due to emf produced due to motion of conducting rod, as a result of which the free electrons in the rod experiences a force


Therefore, electric field needed to balance magnetic force is 0.01Vm-1


(c) We know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by



Since v⃗ and B⃗ are perpendicular and their cross product is parallel to l⃗, eqn.(i) reduces to



Putting the values of v, B and l we get,



Therefore motional emf between the ends of the rod is


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