Question 222: Electromagnetic Induction

Question

Initially a rectangular coil with length vertical is moving out with constant velocity ' $v$ ' in a constant magnetic field ' $B$ ' towards right. Now the same coil is rotated through $90^{\circ}$ in same plane in same magnetic field B and the coil is moving with same velocity $\mathbf{v}$. The magnitude of induced e.m.f. is now

Accepted Answer

Magnetic Flux
: The magnetic flux, written as $$\phi$$, shows how much magnetic field goes through the coil. It is given by $$ \phi = B A \cos 	heta $$, where:
$$B$$ is the strength of the magnetic field,
$$A$$ is the area of the coil,
$$	heta$$ is the angle between the magnetic field and the normal (an imaginary line perpendicular) to the coil's area.
Induced e.m.f
: The e.m.f (voltage) induced in the coil is found by how fast the magnetic flux changes. This is written as:

$$	ext{Induced emf} = -\frac{d\phi}{dt}$$
If the coil is spinning or changing its angle in the magnetic field, the formula becomes:

$$ 	ext{Induced emf} = B A \omega \sin 	heta $$

where $$\omega$$ is the rate at which the coil is being rotated.
At $$	heta = 90^{\circ}$$, $$\sin 90^{\circ} = 1$$, so the induced e.m.f is at its highest value.
After the coil is turned another $$90^{\circ}$$, $$	heta = 180^{\circ}$$. At this angle, the direction between the coil's normal and the magnetic field has changed even more.
So, after rotating the coil by $$90^{\circ}$$, the size of the induced e.m.f is now bigger than what it was at the start.

Answer

greater than initial induced e.m.f.

Answer

less than initial induced e.m.f.

Answer

equal to initial induced e.m.f.

Answer

sometimes greater and sometimes less than initial induced e.m.f.

Electromagnetic Induction

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