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Byju's Answer

Standard XII

Physics

Moment of Inertia of a Rod

A circular wi...

Question

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 $z = \sqrt{3} R$ along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of $45^{\circ}$ with respect to the z-axis. lf the mutual inductance between the loops is given by $\frac{μ_{0} a^{2}}{2^{p / 2} R}$ , then the value of $p$ is

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Solution

The correct option is A $7$
$M = \frac{N ϕ}{I}$
$= \frac{2 \frac{μ_{o} I R^{2}}{2 (8 R^{3})} a^{2} cos 45^{o}}{I}$
$= \frac{μ_{o} a^{2}}{8 R \times 2^{1 / 2}} = \frac{μ_{o} a^{2}}{R \times 2^{7 / 2}}$
$S o,$ $p = 7$

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The correct option is A 7M=NϕI =2μoIR22(8R3)a2cos45oI=μoa28R×21/2=μoa2R×27/2So, p=7

The correct option is A $7$
$M = \frac{N ϕ}{I}$
$= \frac{2 \frac{μ_{o} I R^{2}}{2 (8 R^{3})} a^{2} cos 45^{o}}{I}$
$= \frac{μ_{o} a^{2}}{8 R \times 2^{1 / 2}} = \frac{μ_{o} a^{2}}{R \times 2^{7 / 2}}$
$S o,$ $p = 7$