Apentix/AP Physics 2/Unit 5: Magnetism and Electromagnetic Induction/Question
AP Physics 2easymcq1 pt

A conducting rod of length 0.50 m moves with a constant velocity of 2.0 m/s perpendicular to a uniform magnetic field of 0.40 T. What is the magnitude of the induced emf between the ends of the rod?

A.B) 0.40 V
B.C) 0.80 V
C.A) 0.20 V
D.D) 1.0 V

Explanation

Core Concept

When a conductor of length L moves perpendicular to a magnetic field with velocity v, the induced emf is given by ε = BLv. Therefore, ε = (0.40 T)(0.50 m)(2.0 m/s) = 0.40 V.

Correct Answer

AB) 0.40 V

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