JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 15)
Two insulated circular loop A and B of radius '$$a$$' carrying a current of '$$\mathrm{I}$$' in the anti clockwise direction as shown in the figure. The magnitude of the magnetic induction at the centre will be :
_30th_January_Morning_Shift_en_15_1.png)
$$\frac{\sqrt{2} \mu_0 I}{a}$$
$$\frac{\mu_0 I}{\sqrt{2} a}$$
$$\frac{\mu_0 \mathrm{I}}{2 \mathrm{a}}$$
$$\frac{2 \mu_0 I}{a}$$
Explanation
_30th_January_Morning_Shift_en_15_2.png)
$$\because B_{n e t}=\frac{\sqrt{2} \mu_0 I}{2 a}$$
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