A metal rod of length 2.0 m is moved at 6.0 m/s in a direction perpendicular to its length.A 5.0 T magnetic field is perpendicular to both the rod and its velocity.What is the magnitude of the electric field in the rod?
A) 12 mV/m
B) 60 mV/m
C) 30 mV/m
D) 15 mV/m
E) 0.0

Question

Quizplus · Accepted Answer

When a conductor moves perpendicular to a magnetic field, a voltage is induced across its ends. This is known as the electromagnetic induction phenomenon. The magnitude of the induced voltage, in volts per meter, is given by the formula:

E = Bvl

where E is the electric field, B is the magnetic field, v is the velocity of the conductor, and l is the length of the conductor.

In this problem, the velocity and the magnetic field are given, and the length of the conductor is 2 meters. Therefore, we can calculate the electric field as:

E = (5.0 T) x (6.0 m/s) x (2.0 m) = 60 V/m

However, the problem asks for the magnitude of the electric field, which is given by the absolute value of E. Also, the answer should be in millivolts per meter, not volts per meter. Therefore, we have:

|E| = 60 x 10^3 mV/m = 30 mV/m

Hence, the correct choice is (C).

A Metal Rod of Length 2