In the figure, a conducting rod of length L = 30.0 cm moves in a magnetic field

Question

In the figure, a conducting rod of length L = 30.0 cm moves in a magnetic field B of magnitude 0.440 T directed into the plane of the figure. The rod moves with speed v = 7.00 m/s in the direction shown.

A)When the charges in the rod are in equilibrium, what is the magnitude E of the electric field within the rod? Express your answer in volts per meter to at least three significant figures.

B)What is the magnitude Vba of the potential difference between the ends of the rod? Express your answer in volts to at least three significant figures.

C) What is the magnitude E of the motional emf induced in the rod? Express your answer in volts to at least three significant figures.

Explanation / Answer

When rod is moving the charges within rod experience magnetic field given by Fb = qvB and if E is electric field in rod then force due to electric field is given by Fe = qE. In equilibrium these force will be equal

qvB = qE

E = vB = 7*.44 = 3.08 V/m

Part B) EMf is given by V = E*L ( Here is lenght of rod) = 3.08*0.3 = 0.924 V

Part C) Magnitude of motional emf is same as induced emf in the rod which is 0.924 V

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