0
$\begingroup$

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.

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ The answer is yes to an emf being induced as you do not need a complete conducting circuit to induced the emf. $\endgroup$ – Farcher Feb 23 '18 at 9:28
  • $\begingroup$ @Farcher But isn't emf induced only when there is a change in flux? If the object is moving inside a uniform magnetic field, then the number of field lines passing through the conductor is always the same until the instant it just leaves the field. Why then should an emf be induced at all, regardless of whether it is a complete conducting circuit or not? $\endgroup$ – Archee De Feb 23 '18 at 11:20
  • $\begingroup$ Your question is probably a duplicate. If you put in motional emf into this site's search engine you will find a number of answers relating to induced emf and motional emf. Here is one to read physics.stackexchange.com/q/239741 $\endgroup$ – Farcher Feb 23 '18 at 12:13
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'd add that for motional emf, the seat of the emf is the magnetic Lorentz force, as explained in the above answer. $\mathscr E=-\frac{d\Phi}{dt}$ is just a convenient way of calculating the motional emf. Its advantage is that it automatically sums the emfs induced in different parts of the circuit that may be moving. $\endgroup$ – Philip Wood Jun 28 '19 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.