A straight wire carrying a current of 14 ~A is bent into a semi-circular arc of radius 2.2 ~cm as shown in the figure. The magnetic field produced by the current at the centre (O) of the arc. is \underline\hspace2cm \times ~10^-4 ~T

Magnetics PYQ — JEE Main 2023 · 10 Apr · Shift 2 · Q58 with Solution

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Correct answer: 2

For a circular arc of angle $\theta$ (in radians), radius $R$ , carrying current $I$ , the magnetic field at the centre is

B = \frac{\mu_0 I \theta}{4\pi R}

For a semi-circle, $\theta = \pi$ . Hence,

B = \frac{\mu_0 I}{4R}

Given:

So,

B = \frac{4\pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}}

Cancel $4$ :

B = \frac{\pi \times 14 \times 10^{-7}}{2.2 \times 10^{-2}}

B = \frac{14\pi}{2.2} \times 10^{-5}

Now,

\frac{14}{2.2} \approx 6.36

Thus,

B \approx 6.36\pi \times 10^{-5}

Using $\pi \approx 3.14$ ,

B \approx 19.98 \times 10^{-5} = 1.998 \times 10^{-4}\,\text{T}

B \approx 2 \times 10^{-4}\,\text{T}

Since the field is of the form

\underline{\hspace{2cm}} \times 10^{-4}\,\text{T}

the required integer is

\boxed{2}

Stored correct answer = $2$

Our derived answer also = $2$ . Hence, they agree.