Courses for Kids
Free study material
Offline Centres

A current carrying wire is in the form of square loop of side length$10cm$. Current wire is $5A$ . Find the magnetic field at the center of the loop.
(A) ${10^{ - 7}}T$
(B) $\sqrt 2 \times {10^{ - 7}}T$
(C) $4\sqrt 2 \times {10^{ - 5}}T$
(D) $4 \times {10^{ - 7}}T$

Last updated date: 01st Mar 2024
Total views: 21k
Views today: 1.21k
Rank Predictor
21k+ views
Hint We are given the side of the current carrying square loop and the current flowing through it and are asked to find the magnetic field at the center of the loop. We will use the concepts where we could connect the current at a point to the magnetic field and the length of the loop.
Formula Used
$B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{I}{a}(\cos {\theta _1} + \cos {\theta _2})$
Where, $B$ is the magnetic field at a point, ${\mu _0}$ is the permeability of open space or vacuum, $I$ is the current flowing through the conductor and $a$ is the distance of the point from the edge of the conductor. ${\theta _1}$ and ${\theta _2}$ are the angle between the line joining the point with one edge and the edge of the loop.

Complete Step By Step Solution

We have constructed a line $OA$ to the mid of one of the edges of the square.
As $O$ is the center of the square, the angles ${\theta _1} = {\theta _2} = \dfrac{\pi }{4}$ (As edge angle of a square is $\dfrac{\pi }{2}$)
The parameters given are,
Current through the conductor, $I = 5A$
Length of each edge, $L = 10cm = {10^{ - 1}}m$
As per the diagram,
Distance of the edge of the square from the center, $a = 0.5 \times {10^{ - 1}}m = 5 \times {10^{ - 2}}m$
And we know,
${\mu _0} = 4\pi \times {10^{ - 7}}H{m^{ - 1}}$
Thus, applying the formula $B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{I}{a}$ , we get
\[B = \dfrac{{4\pi \times {{10}^{ - 7}}H{m^{ - 1}}}}{{4\pi }} \times \dfrac{{5A}}{{5 \times {{10}^{ - 2}}m}}\left( {\cos \dfrac{\pi }{4} + \cos \dfrac{\pi }{4}} \right)\]
Calculating and putting in the value $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$, we get
$B = \sqrt 2 \times {10^{ - 5}}T$
Now, each and every edge will apply the same magnetic field on the center
\[Net{\text{ }}magnetic{\text{ }}field{\text{ }}on{\text{ }}the{\text{ }}center{\text{ }} = {\text{ }}4{\text{ }} \times magnetic{\text{ }}field{\text{ }}by{\text{ }}each{\text{ }}edge = {\text{ }}4 \times B\]
${B_{Net}} = 4\sqrt 2 \times {10^{ - 5}}T$

Thus, the answer is (C).

Note We multiplied the number of edges with the individual magnetic field of each edge in order to get the net magnetic field on the center. But this was the case of a square where each and every edge has the same dimension. If the loop was of a different shape, then the calculations would have been somewhat different from the one we did. As this loop was a square, it was safe to assume that each edge will apply the same amount of magnetic field on the center.