The current of 5 A flows in a square loop of sides 1 m placed in air. The magnetic field at the center of the loop is X 2 10^-7 T. The value of X is:

Given: Current in the square loop is I = 5 A and side length is L = 1 m. The magnetic field at the center is written as X 2 10^-7 T. Find: The value of X. Use the magnetic field due to one finite straight side of the square and then add the contribution of all four sides. For one side, B = 0 I4 d( 1 + 2) At the center of the square, d = L2 = 12 = 0.5 m and 1 = 2 = 45^ So, the magnetic field due to one side is B one side = 0 I4 d( 45^ + 45^) Substituting the values, B one side = (4 10^-7) 54 0.5 ( 1 2 + 1 2) B one side = 5 10^-70.5 ( 2 2) B one side = 10 10^-7 2 = 2 10^-6 T The total magnetic field at the center is four times this value: B center = 4 B one side B center = 4 ( 2 10^-6) = 4 2 10^-6 T Rewrite it in the required form: B center = 40 2 10^-7 T Comparing with X 2 10^-7 = 40 2 10^-7 we get X = 40 Therefore, the correct option is D.

The current of 5 A flows in a square loop of sides 1 m placed in… — Solution | JEE Physics

MCQMediumJEE 2024Biot–Savart Law

The current of $5 \, \text{A}$ flows in a square loop of sides $1 \, \text{m}$ placed in air. The magnetic field at the center of the loop is $X\sqrt{2} \times 10^{-7} \, \text{T}$ . The value of $X$ is:

A
$10$
B
$20$
C
$30$
D
$40$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Current in the square loop is $I = 5 \, \text{A}$ and side length is $L = 1 \, \text{m}$ . The magnetic field at the center is written as $X\sqrt{2} \times 10^{-7} \, \text{T}$ .

Find: The value of $X$ .

Use the magnetic field due to one finite straight side of the square and then add the contribution of all four sides.

For one side,

B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)

At the center of the square,

d = \frac{L}{2} = \frac{1}{2} = 0.5 \, \text{m}

and

\theta_1 = \theta_2 = 45^\circ

So, the magnetic field due to one side is

B_{\text{one side}} = \frac{\mu_0 I}{4\pi d}(\sin 45^\circ + \sin 45^\circ)

Substituting the values,

B_{\text{one side}} = \frac{(4\pi \times 10^{-7}) \times 5}{4\pi \times 0.5}\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)

B_{\text{one side}} = \frac{5 \times 10^{-7}}{0.5}\left(\frac{2}{\sqrt{2}}\right)

B_{\text{one side}} = 10 \times 10^{-7} \times \sqrt{2} = \sqrt{2} \times 10^{-6} \, \text{T}

The total magnetic field at the center is four times this value:

B_{\text{center}} = 4 \times B_{\text{one side}}

B_{\text{center}} = 4 \times (\sqrt{2} \times 10^{-6}) = 4\sqrt{2} \times 10^{-6} \, \text{T}

Rewrite it in the required form:

B_{\text{center}} = 40\sqrt{2} \times 10^{-7} \, \text{T}

Comparing with

X\sqrt{2} \times 10^{-7} = 40\sqrt{2} \times 10^{-7}

we get

X = 40

Therefore, the correct option is D.

Common mistakes

Using the distance from the center to a vertex instead of the perpendicular distance from the center to a side. This is wrong because the finite-wire formula requires the perpendicular distance $d$ from the point to the wire. Use $d = \frac{L}{2}$ , not the half-diagonal.
Taking the angle for each side as $90^\circ$ instead of $45^\circ$ . This is wrong because the lines from the center to the two ends of a side make equal angles of $45^\circ$ with the perpendicular to that side. Use $\sin 45^\circ + \sin 45^\circ$ .
Forgetting to add the contribution of all four sides. This is wrong because by symmetry each side produces the same magnetic field at the center and all are in the same direction. First find the field due to one side, then multiply by $4$ .

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