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.0.20 V
B.0.40 V
C.0.80 V
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

B0.40 V

More Unit 5: Magnetism and Electromagnetic Induction practice questions

Try a random question →

Practice more AP Physics 2 questions with full explanations

Practice Unit 5: Magnetism and Electromagnetic Induction Questions →