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Does emf get induced for any conductor moving perpendicularly to a magnetic field?

In the textbook explanation, they gave an example of a fixed metal frame over which a metal rod can roll. The field goes into the area of the frame and the rod moves horizontally over the frame, such that its motion is perpendicular to the field. In this case, the area bounded by the frame and the rod keeps changing, hence the flux changes, and an emf is induced.

But consider a case where there is no frame, and the rod is just moving perpendicular to a magnetic field with constant velocity. Is emf still induced across the ends of the rod?

Also, is the shape of the conductor of any significance? For example, replace the straight rod with a semicircular one in the above question.

Archee De
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1 Answers1

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Induced emf is not defined w.r.t change in flux. It is defined as,

$$\xi=\int \vec{f}_{mag}.\vec{dl}$$

where where $f_{mag}$ is the total force per unit charge which drives the current around the circuit. This can be due to the battery or can be due to a non electrostatic electric field or a magnetic field. The source cannot be an electrostatic field since the line integral of an electrostatic field over the entire circuit is $0$ since it is a conservative field.

Now if there is a single rod moving in a region with perpendicular magnetic field, there will be seperation of charges due to the Lorentz force, $\vec{f}_{mag}=\vec{v} \times\vec{B}$. Thus an emf will be induced according to the previous formula. Since the circuit is not complete, there won't be any current flow.

For your second question, the answer is a clear no. Induced emf only depends on the line integral. It can be calculated for any kind of shape you can come up with.

If you still have doubts regarding induced emf, you can refer to the book by David J. Griffiths.

Satwata Hans
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