JEE Main 2022 · 28 Jul · Shift 2 · Physics Q13

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PhysicsSingle correct+4 / -1Question 13 of 90

The magnetic field at the center of current carrying circular loop is $B_1$ . The magnetic field at a distance of $\sqrt3$ times radius of the given circular loop from the center on its axis is $B_2$ . The value of $B_1/B_2$ will be

9 : 4

12:\sqrt5

8 : 1

5:\sqrt3

Answer and worked solution

Official answer: C

On-axis field. $\frac{B_1}{B_2}=\left(1+\frac{x^2}{R^2}\right)^{3/2}=4^{3/2}=8$ .

(C)

Topics:Biot-Savart law

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JEE Main 2022 · 28 Jul · Shift 2, Physics question 13