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: Find the magnetic field inside the solenoid.
For a long solenoid with $N$ turns per unit length carrying current $I = I_0 \sin(\omega t)$, the magnetic field inside is: \[ B = \mu_0 N I = \mu_0 N I_0 \sin(\omega t) \] The field is uniform inside the solenoid and directed along its axis.
Step 2: Find the area of the square loop.
The square loop has its corners touching the circular cross-section of the solenoid (radius $R$). This means the diagonal of the square equals the diameter of the circle: \[ \text{diagonal} = 2R \] If the side length of the square is $a$, then by the Pythagorean theorem: \[ a\sqrt{2} = 2R \implies a = \sqrt{2}R \] Area of the square: \[ A = a^2 = (\sqrt{2}R)^2 = 2R^2 \]
Step 3: Calculate the magnetic flux through the square loop.
The plane of the square loop is perpendicular to the solenoid axis, so the field is perpendicular to the loop (i.e., $\cos 0^\circ = 1$). The flux is: \[ \Phi = B \cdot A = \mu_0 N I_0 \sin(\omega t) \times 2R^2 = 2\mu_0 N I_0 R^2 \sin(\omega t) \]
Step 4: Apply Faraday's law of electromagnetic induction.
The induced EMF is: \[ e = -\frac{d\Phi}{dt} \] Taking the time derivative of the flux: \[ \frac{d\Phi}{dt} = 2\mu_0 N I_0 R^2 \omega \cos(\omega t) \]
Step 5: Write the magnitude of induced EMF.
\[ |e| = 2\mu_0 N I_0 R^2 \omega \cos(\omega t) \] The negative sign from Faraday's law indicates the direction (Lenz's law), but the magnitude of the EMF is as above.
Step 6: State the final answer.
The EMF induced in the square coil is: \[ \boxed{2\mu_0 N I_0 R^2 \omega \cos(\omega t)} \]