Question

When a wire loop is rotated in a magnetic field, the direction of the induced emf changes once in each
$(A)$ Changes every $\dfrac{1}{4}$ revolution’
$(B)$Does not change
$(C)$Change every $\dfrac{1}{2}$ revolution
$(D)$Changes every $1$ revolution.

Answer 1

Hint: The magnetic flux is described as several magnetic field lines passing through the closed surface. It helps to measure the magnetic field. Weber is the SI unit of the magnetic flux. To find the solution we have to use the different angles for finding the revolution.

Formula Used:
The formula used to solve the above problem as follows,
The flux of the magnetic field through the loop is as follows,
\[\phi = B.Acos\omega t\]

Complete step by step solution:
We know that the flux of the magnetic field through the loop is given by,
\[\phi = B.Acos\omega t\]
We know E.M.F is given by,
\[EMF = \dfrac{{ - d\phi }}{{dt}}\]
Now substitute the values in the above equation we get,\[EMF = \dfrac{{ - d(B.A\cos \omega t)}}{{dt}}\]
Now differentiate the above equation, then it becomes as follows
\[EMF = B.A\sin \omega t\]
To find the revolution we have to find the e.m.f for different angles,
The given angles are, \[\omega t = 0to\pi \]
\[\sin \omega t = + ve\]
For the given angles, \[\omega t = \pi to2\pi \]
\[\sin \omega t = - ve\]
The direction of the emf changes after changing the \[\pi \] angle, hence this \[\pi \] angle represents the half revolution.
Finally, the direction of the emf changes after the half revolution
Here the option $(C)$ is the correct answer.

Note:
The magnetic field at each part of the rotating loop is determined by the rule of thumb on the right side. Usually, the magnetic field lines are parallel at the revolving circle in the middle. The magnetic field is a vector field that defines the magnetic effect in magnetic materials.