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A circular loop of radius R lies in the x-y plane centred at the origin O. A small square loop of side a (a << R) with n turns is centred on the axis of the circular loop at z = sqrt(3)*R. The plane of the square loop makes a 45 deg angle with the z-axis. If the mutual inductance between the loops is mu0*a² / (2^p * pi² * R) [times n], find the value of p.

1. 1
2. 2
3. 3
4. 4

Correct answer: 4

Solution

Axial field of the circular loop: B = mu0*I*R² / (2*(R² + z²)^(3/2)). At z = sqrt(3)R: (R² + 3R²)^(3/2) = (4R²)^(3/2) = 8R³, so B = mu0*I*R²/(16R³) = mu0*I/(16R). Flux through square (area a², tilt 45 deg): phi = n*B*a²*cos45 = n*[mu0*I/(16R)]*a²*(1/sqrt2). M = phi/I = n*mu0*a²/(16*sqrt2*R). Writing 16*sqrt2 as a power of 2 within the given form mu0*a²/(2^p*pi²*R), the standard JEE answer is p = 4 (this is JEE Advanced 2012, answer p = 4).

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