Question

A solenoid 4cm in diameter and 20cm in length has 250 turns and carries a current of 15A. Calculate the flux through the surface of a disc 10cm radius that is positioned perpendicular to and centered on the axis of the solenoid.

Answer 1

The magnetic field inside a solenoid is given by the formula $B = {\mu _o}ni$, where n denotes the number of turns per unit length and i denotes the current in the solenoid. Flux through any area A having magnetic field B perpendicular to it can be calculated using the formula $\phi = BA$. It should be remembered that area A here denotes the area through which the magnetic field is crossing the surface and not the total area of the surface.

Complete step by step answer:
The magnetic field inside a solenoid is given by $B = {\mu _o}ni$
Here it is given that the total number of turns of the solenoid is 250 and its length is 0.2m.
Hence, we can get the number of turns per unit length using $n = \dfrac{{250}}{{0.2}}$,
Therefore, $n = 1250$.
Current in the solenoid is given and it is 15A, hence $i = 15A$.
The value of permeability of vacuum is given by ${\mu _o} = 4\pi \times {10^{ - 7}}$.
Putting all these values in the formula of magnetic field, we get
$B = 4\pi \times {10^{ - 7}} \times 1250 \times 15$,
$B = \dfrac{{165}}{{28}} \times {10^{ - 3}}$,
Now, the area through which this magnetic field is crossing is only the area of the solenoid, since the magnetic field outside the solenoid is assumed to be zero.
Cross section area of the solenoid having 2cm radius will be $A = \pi {r^2}$,
$A = \pi \times {0.02^2}$,
$A = \dfrac{\pi }{{2500}}$
Hence the flux passing through the disc will be
$\phi = BA$,
$\phi = \dfrac{{165}}{{28}} \times {10^{ - 3}} \times \dfrac{\pi }{{2500}}$
$\phi = 7.4 \times {10^{ - 6}}$Weber.

Note: Before attempting to solve the problem, the student needs to be able to understand that the solenoid has a field only inside it and the net field outside the solenoid is zero. One should also understand that it does not matter what the area of the surface is through which flux needs to be calculated, during calculation of flux we only use the area through which magnetic field lines are crossing. Like here, we had a large disc but the magnetic field was only crossing through 4cm diameter i.e. the part inside the solenoid so we used only that area for calculation of flux.