JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Electromagnetic Induction

A circular wire loop of radius

R

is placed in the

x - y

plane centered at the origin

O

. A square loop of side

a (a ≪ R)

having two turns is placed with its center at

z = 3 R

along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of

4 5^{\circ}

with respect to the

z

-axis. If the mutual inductance between the loops is given by

\frac{μ _{0} a ^{2}}{2 ^{p /2} R}

, then the value of

p

Answer: 7

Solution

The magnetic field at the center of the square loop due to the circular loop is

B = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + 3 R ^{2} ) ^{3/2}} = \frac{μ _{0} I R ^{2}}{2 ( 4 R ^{2} ) ^{3/2}} = \frac{μ _{0} I}{16 R}

. The flux through the square loop (2 turns, area

a^{2}

, angle

4 5^{\circ}

) is

Φ = 2 \times B \times a^{2} cos 4 5^{\circ} = 2 \times \frac{μ _{0} I}{16 R} \times a^{2} \times \frac{1}{2} = \frac{μ _{0} I a ^{2}}{8 R 2}

. The mutual inductance is

M = \frac{Φ}{I} = \frac{μ _{0} a ^{2}}{8 R 2} = \frac{μ _{0} a ^{2}}{R \cdot 2 ^{7/2}}

. Comparing with

\frac{μ _{0} a ^{2}}{2 ^{p /2} R}

, we get

p = 7

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Integer answer