Question

The phase difference between the flux linked with a coil rotating in a uniform magnetic field and induced e.m.f. produced in it is-
(A) $\dfrac{\pi }{2}$
(B) $\dfrac{\pi }{3}$
(C) $\dfrac{\pi }{6}$
(D) $\pi $

Answer 1

Hint: The value of induced e.m.f. produced in the circuit is given by rate of change of flux passing through the coil. This value can be related to the flux linked with a coil in rotating in a magnetic field, and the trigonometric relation between sin and cos can be used to determine the phase difference between both quantities.
Formula used
$\varepsilon = - \dfrac{{d\phi }}{{dt}}$
$\phi = BA\cos \theta $
Where $\varepsilon $is the induced e.m.f.
B is the magnetic field
A is the area
$\theta $is the angle between area vector and magnetic field lines
$\phi $ is the magnetic flux passing through the coil.

Complete step by step solution:
From the lenz law, we know that the e.m.f. produced by a moving coil in a uniform magnetic field is given by the formula
$\varepsilon = - \dfrac{{d\phi }}{{dt}}$
Where $\varepsilon $represents the e.m.f. produced and $\phi $ is the amount of flux passing through the coil at a given instant.
The minus sign represents that the direction of the produced e.m.f. is such that it opposes it’s cause.
The value of flux is calculated by the formula-
$\phi = BA\cos \theta $
Here B is the value of the magnetic field, A is the area enclosed by the loop, and $\theta $ is the angle created between the area vector and the magnetic field.
Now we can replace the value of the $\phi $in the expression of e.m.f. produced,$\varepsilon $.
$\varepsilon = - \dfrac{d}{{dt}}\left( {BA\cos \theta } \right)$
On differentiating the expression, we get-
$\varepsilon = BA\sin \theta $ ($\dfrac{d}{{dx}}\cos x = - \sin x$)

As per trigonometric relation between $\sin \theta $and $\cos \theta $, we know that
$\cos \left( {\dfrac{\pi }{2} + x} \right) = \sin x$
Using this relation in the expression of $\varepsilon $, we have-
$\varepsilon = BA\cos \left( {\dfrac{\pi }{2} + \theta } \right)$
On comparing this term with the value of $\varepsilon $ from the Lenz law we can conclude that there is a phase difference which is equal to $\dfrac{\pi }{2}$.

Thus option (A) is correct.

Note: In the calculations where magnetic flux is involved, the areas are treated as vector quantities. The normal which extrudes out of the plane area is treated as the direction of the area vector. For the flux, the area in front of the field is given by the cosine of the original area.