Question

A long solenoid is carrying a current \[ I=I_0\sin(\omega t), \] having \(N\) turns per unit length and radius \(R\). A square loop is placed inside the solenoid with its plane perpendicular to the solenoid axis, and its corners touching the solenoid. Now the e.m.f induced in the square coil is:

• A. \(\mu_0NI_0R^2\sin(\omega t)\)
• B. \(2\mu_0NI_0R^2\sin(\omega t)\)
• C. \(2\mu_0NI_0R^2\omega\cos(\omega t)\)
• D. \(\mu_0NI_0R^2\pi\omega\cos(\omega t)\)

Hint:

For a square inscribed inside a circle of radius \(R\), \[ \text{Area of square}=2R^2 \] Also, induced e.m.f. is obtained using \[ e=-\frac{d\Phi}{dt} \] where \(\Phi=BA\).

Answer 1

The Correct Option is C

Step 1: Magnetic field inside the solenoid.
The magnetic field inside a long solenoid is \[ B=\mu_0 n I \] Here, \[ n=N \] (turns per unit length) Given current is \[ I=I_0\sin(\omega t) \] Therefore, \[ B=\mu_0NI_0\sin(\omega t) \]

Step 2: Area of the square loop.
The square is placed inside the solenoid such that all four corners touch the circular cross-section of the solenoid.
Hence, the diagonal of the square equals the diameter of the solenoid: \[ d=2R \] If side of square is \(a\), then \[ a\sqrt{2}=2R \] Thus, \[ a=\frac{2R}{\sqrt{2}} \] \[ a=\sqrt{2}R \] Area of square: \[ A=a^2 \] \[ A=(\sqrt{2}R)^2 \] \[ A=2R^2 \]

Step 3: Magnetic flux through the square loop.
Magnetic flux is \[ \Phi=BA \] Substituting the values, \[ \Phi=\mu_0NI_0\sin(\omega t)\times2R^2 \] \[ \Phi=2\mu_0NI_0R^2\sin(\omega t) \]

Step 4: Calculate the induced e.m.f.
By Faraday’s law, \[ e=-\frac{d\Phi}{dt} \] Therefore, \[ e=-\frac{d}{dt} \left( 2\mu_0NI_0R^2\sin(\omega t) \right) \] \[ e=-2\mu_0NI_0R^2\omega\cos(\omega t) \] Magnitude of induced e.m.f. is \[ e=2\mu_0NI_0R^2\omega\cos(\omega t) \]

Step 5: Final conclusion.
Therefore, the induced e.m.f. is \[ \boxed{2\mu_0NI_0R^2\omega\cos(\omega t)} \]