A small square loop of wire of side l is placed inside a large square loop of wire of side `L(Lgtgtl)`. The loops are co-planer and their centres coincide. The mutual inductance of the system is proportional to

Options

A. `l/L`
B. `l^(2)/L`
C. `L/l`
D. `L^(2)/l`

Answer

B

Explanation

The magnetic field due to a current flowing in a wire of finite length is given by
`B=(mu_0 I)/(4 pi R)(sin alpha+sin beta)`
Appling the above formula AB for finding the field at O, we get
`B=(mu_(0)I_(1))/(4 pi(L//2))(sin 45^@+sin 45^@0=(mu_(0)I_(1))/(sqrt(2)piL)`
Acting perpendicular to the plane of paper upwads
`:. The total magnetic field due to current flowing through ABCD is

`B=4(B_1)=(4 mu_(0)(I_(1))/(sqrt(2)pi L)=(2sqrt(2)mu_(0)I_(1))/(piL)`
The total flux passing through the square EFGH
`(phi_2)=Bxx(l^2)=(2sqrt(2)mu_(0)I_(1))/(piL)xx(l^2)` ...(i)
[`:. l gt L` and therefore, B can be assumed constat for `(l^2)`]
The flux through small square loop is directly proportional to the current passing through big square loop.
`:. (phi_2)(prop)(I_1)implies(phi)=(mI_(1))` where M=Mutual Conductance
`:. (M_2)(phi_2)/(I_1)==((2sqrt(2)mu_(0)I_(1))/(piL)xx(l^2))/(I_(1))`=`(2sqrt(2)mu_(0))/(piL)xx(l^2)`
`impliesm (prop) (l^2)/(L)`.