0
$\begingroup$

heres a diagram of a solenoid and another coil connected to the galvanometer

If the coil on the left is rotated about the magnetic field lines, will there be an induced current? I know that the flux concerns only the component of the magnetic field perpendicular to the plane of the coil, so even if it is rotated, I assume the perpendicular component will remain the same. Just want to confirm this. Thanks

$\endgroup$
0
1
$\begingroup$

The flux is defined as integal {if you don't know calculus, take this as a sum} of $\int_\Gamma \vec B \dot \, \mathrm d \vec A$ where dA points out the surface $\Gamma$.


(source: gsu.edu)

if you had a circular shape of coil, the flux would not change, and hence, no Electric field or EMF would be induced, but here, as the field of solenoid may not be perfectly symmetrical outside it, (spreads out evenly in each direction) . hence, there is bound to be some emf, not too much, theoretically for an ideal solenoid, there would be no emf induced too.

$\endgroup$
1
  • $\begingroup$ Please see this help post to learn how to write your equations in a way nicer way i.e. in $\LaTeX$, in order to improve legibility. I did it for you this time but in the future it'd be better if you did it by yourself. Thanks! $\endgroup$
    – Gonenc
    Feb 12 '16 at 8:14
-1
$\begingroup$

In general, a inductive 'loop' for picking up a magnetic field (as shown connected to the galvanometer in you diagram) is 'symmetric' with respect to rotation about the axis parallel to the magnetic field from the coil connected to the battery. If the 'loop' is square (as shown), the symmetry is perfect for 90 degree rotations. There may be some small variation in the magnetic flux enclosed by a 'square' coil rotated at 45 degrees for example. A circular loop rotated about it's centre would be a better chioce.

$\endgroup$
-1
$\begingroup$

Depends on the angular velocity of its rotation. Electromagnetic interactions propagates at light speed. This means, if the coil is rotating in a frequency low enough, and if the induction circuit is close enough, the magnetic fields will arrive in the other circuit, almost "statically". Therefore, as good approximation in this case, you can consider the magnetic field arriving in the induction circuit, arrived "instantaneously" from the coil.

If coil has length $L$, rotating at angular speed $\omega$, and the induction circuit is at a distance $s$ from the coil, then our approximations: The extreme part of the coil must have speed lower than speed of light: $L\omega\ll c$. And the time for the magnetic field arrive the circuit must be close to zero: $\Delta t = s/c\ll 1$. If this is true, you can calculate the emf with induction law approximating that the magnetic field rotates nicely with the coil: $$ \mbox{emf} = -\frac{d\Phi}{dt} $$

$\endgroup$

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.