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

Question

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

Filo · Accepted Answer

Answer: BSolution: The magnetic field due to a current flowing in a wire of finite length is given by

B = \frac{μ _{0} I}{4 π R} (sin α + sin β)

Appling the above formula AB for finding the field at O, we get

B = \frac{μ _{0} I _{1}}{4 π ( L / 2 )} (sin 45^{\circ} + sin 45^{\circ} 0 = \frac{μ _{0} I _{1}}{2 π L}

Acting perpendicular to the plane of paper upwads

∴ T h e \to t a l m a g \neq = t i c f i e l d d u e \to c u r r e n t f l o w \in g t h r o u g h A B C D i s

B=4(B_1)=(4 mu_(0)(I_(1))/(sqrt(2)pi L)=(2sqrt(2)mu_(0)I_(1))/(piL)\displaystyle(phi_2)=Bxx(l^2)=(2sqrt(2)mu_(0)I_(1))/(piL)xx(l^2)

\dots (i) [

:. l gt L

and ∴, B c a n b e a s \sum e d c o n s t a t f or

(l^2)\displaystyle{]}:. (phi_2)(prop)(I_1)implies(phi)=(mI_(1))

w h e r e M = M u t u a l C o n d u c tan c e

:. (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)\displaystyle.

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Question Text	A small square loop of wire of side l is placed inside a large square loop of wire of side $L (L >> l)$ . The loops are co-planer and their centres coincide. The mutual inductance of the system is proportional to
Answer Type	Text solution:1
Upvotes	150

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

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