Question

A long solenoid with 20 turns per cm has a loop of area 2.0 $cm^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 sec, calculate the induced emf in the loop while the current is changing.

Answer 1

Hint: We know by Faraday's law that a changing current produces changing magnetic fields and this produces a change in magnetic flux. A change in magnetic flux will result in an induced e.m.f. which will oppose the changing flux (will have a minus sign).

Formula used:
The magnitude of the emf produced will be given by:
$e = \dfrac{d \phi}{dt}$
where $\phi$ is magnetic flux.

Complete answer:
A solenoid consists of a piece of conducting wire (copper) turned into a number of coils (closely placed). Solenoid is the best for producing magnetic fields with the help of current. The magnetic field produced at the centre of the solenoid is uniform and is given by:
$B = \dfrac{\mu_0 N I}{l}$
In the question, we are already given N/l = 20 turns per cm or 2000 turns per meter.
When the current is 2 A, magnetic field will be
$B_1 = \mu_0 .(2000). 2 T$
When the current is 4 A, magnetic field will be
$B_2 = \mu_0 .(2000). 4 T$
The difference between the two is
$B_2 - B_1 = \mu_0 (4000) T$

Now, we know that magnetic flux is written as:
$\phi = B.A$
Where $ A= 0.2 cm^2$ is given.

Therefore we obtain
$\dfrac{d \phi}{dt} = \dfrac{0.2 \times 10^{-4} m^2 \times \mu_0 (4000) T }{0.1 s}$
Which is nothing but the magnitude of emf induced. Therefore, we obtain:
$e = 1.0053 \times 10^{-6}$ V or 1$\mu$V.

Additional Information:
Solenoids are primarily used in research and industries to produce a uniform magnetic field. Magnetic field lines pass parallel to the axis inside a solenoid. If we use permanent magnets, we know that we observe curved field lines. Solenoids are also special because their magnetism can be turned on or off with the help of an external supply while this is never achievable if we were using permanent magnets.

Note:
The formula for flux contains an angle dependent term. If the loop of area 0.2 $cm^2$ was making an angle with (solenoid) the direction of the magnetic field, then we would have to use the formula BA$\cos \theta$ for magnetic flux.