A small square loop of side a and one turn is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The magnetic field $B$ at the center due to a square loop of side $b$ carrying current $I$ is given by the sum of the fields due to its four sides

Vedclass · Accepted Answer

$\frac{\mu_{0}}{4 \pi} 8 \sqrt{2} \frac{a^{2}}{b}$

Vedclass · Answer

(A) The magnetic field $B$ at the center due to a square loop of side $b$ carrying current $I$ is given by the sum of the fields due to its four sides.For one side,the field at the center is $B_{1} = \frac{\mu_{0} I}{4 \pi (b/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_{0} I}{2 \pi b} (2 	imes \frac{1}{\sqrt{2}}) = \frac{\sqrt{2} \mu_{0} I}{\pi b}$.Since there are four sides,the total magnetic field at the center is $B = 4 	imes B_{1} = \frac{4 \sqrt{2} \mu_{0} I}{\pi b}$.Since $b \gg a$,we assume the magnetic field is uniform over the area of the smaller loop.The magnetic flux $\phi$ linked with the smaller loop of area $A = a^{2}$ is $\phi = B 	imes A = \frac{4 \sqrt{2} \mu_{0} I}{\pi b} 	imes a^{2}$.The coefficient of mutual inductance $M$ is defined as $M = \frac{\phi}{I} = \frac{4 \sqrt{2} \mu_{0} a^{2}}{\pi b}$.To match the given options,we multiply and divide by $4$:$M = \frac{\mu_{0}}{4 \pi} 	imes 16 \sqrt{2} \frac{a^{2}}{b}$.Wait,re-evaluating the field calculation: $B_{side} = \frac{\mu_{0} I}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$. Here $d = b/2$,$	heta_1 = 	heta_2 = 45^{\circ}$.$B_{side} = \frac{\mu_{0} I}{4 \pi (b/2)} (2 \sin 45^{\circ}) = \frac{\mu_{0} I}{2 \pi b} \sqrt{2} = \frac{\mu_{0} I}{\sqrt{2} \pi b}$.Total $B = 4 	imes \frac{\mu_{0} I}{\sqrt{2} \pi b} = \frac{2 \sqrt{2} \mu_{0} I}{\pi b}$.Flux $\phi = B a^{2} = \frac{2 \sqrt{2} \mu_{0} I a^{2}}{\pi b}$.$M = \frac{\phi}{I} = \frac{2 \sqrt{2} \mu_{0} a^{2}}{\pi b} = \frac{\mu_{0}}{4 \pi} \frac{8 \sqrt{2} a^{2}}{b}$.

Similar Questions

Two coils of self-inductance $100\,mH$ and $400\,mH$ are placed very close to each other. Find the maximum mutual inductance between the two. (The current value is irrelevant to the calculation of mutual inductance).

$A$ small circular loop of wire of radius $a$ is located at the centre of a much larger circular wire loop of radius $b$. The two loops are in the same plane. The outer loop of radius $b$ carries an alternating current $I = I_0 \cos (\omega t)$. The emf induced in the smaller inner loop is nearly

$A$ pair of adjacent coils has a mutual inductance of $1.5 \ H$. If the current in one coil changes from $0$ to $20 \ A$ in $0.5 \ s$,what is the change of flux linkage with the other coil (in $Wb$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module