Question

A 1600 turn solenoid is 1.9m long and 15cm in diameter. The solenoid current is increasing at 1kA/s. Find the current in a 23cm diameter loop with resistance 10 ohms lying entirely outside of the solenoid.

Answer 1

Faraday's law ,emf,V developed in a closed loop is given

\(\huge{|V|=\frac{d\phi_{b}}{dt}}\)
\(\large{\phi_{b}}\) is the magnetic flux associated with a surface bounding the closed loop.Then current in the loop is given by Ohm's law I=V/R where R is the resistance of the loop. Here the closed loop will be the 23cm diameter ring.

Try it now. If u are unsuccessful, ask me to solve it.

Answer 2

I tried to solve the problem and was unsuccessful could you solve it for me

Answer 3

Since we need rate of change of magnetic flux through the loop, so let us first find magnetic flux through it at any moment in time when say "i" current is flowing in the solenoid.
Now see the figure. The loop is the bigger circle of 23 cm diameter and the smaller circle is the solenoid. The solenoid is penetrating the plane of the loop. Just imagine a stick poking through a circular ring.

Now comes an important thing. Magnetic field is non zero only inside the solenoid. Outside the solenoid magnetic field is zero. So ,we may break the area of the loop into 2 parts : A1 and A2. So B is 0 in A1 and non zero in A2.

You should know that if "i" current is going through solenoid, then B inside the solenoid is given by ,\(B=\mu_{0} ni\) where n is turns per unit length.

Now to find magnetic flux over the entire area of the loop.
\(\phi_{b}=B_{1}A_{1}+B_{2}A_{2}\)
Here B1 is 0. \(B_{2}=\mu_{0} ni\) , \(A_{2} =\pi r^{2}\) where r is the radius of the loop with area A2 which is 15/2 =7.5cm

So,
\(\phi_{b} =\mu_{0} ni\pi r^{2}\)
\(d\phi_{b}/dt=\mu_{0} n(di/dt)\pi r^{2}\)

Now put the all the values in the above expression and find the value of left hand side which by Faraday is emf in the loop.

<Please ask if u do not understand or see some mistake>