JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 8)
A circular loop of radius $$r$$ is carrying current I A. The ratio of magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is :
3$$\sqrt2$$ : 2
1 : 3$$\sqrt2$$
2$$\sqrt2$$ : 1
1 : $$\sqrt2$$
Explanation
_24th_January_Morning_Shift_en_8_1.png)
$$
\begin{aligned}
B_{P_{1}} & =\frac{\mu_{0} l}{2 r} \\\\
B_{P_{2}} & =\frac{\mu_{0} l r^{2}}{2\left(r^{2}+r^{2}\right)^{3 / 2}}=\frac{\mu_{0} I}{2^{5 / 2} r}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \quad \frac{B_{P_{1}}}{B_{P_{2}}}=\frac{\frac{\mu_{0} I}{2 r}}{\frac{\mu_{0} I}{2^{5 / 2} r}}=\frac{2 \sqrt{2}}{1}
\end{aligned}
$$
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