JEE Main 2022 · 24 Jun · Shift 1 · Physics Q49

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PhysicsSingle correct+4 / -1Question 49 of 90

The magnetic field at the centre of a circular coil of radius $r$ , due to current $I$ flowing through it, is $B$ . The magnetic field at a point along the axis at a distance $\dfrac r2$ from the centre is :

B/2

2B

\left(\dfrac2{\sqrt5}\right)^3B

\left(\dfrac2{\sqrt3}\right)^3B

Answer and worked solution

Official answer: C

Axial field. $B_x=\dfrac{\mu_0Ir^2}{2(r^2+x^2)^{3/2}}=B\left(\dfrac{r^2}{r^2+x^2}\right)^{3/2}$ .

At $x=r/2$ . $B_x=B\left(\dfrac{1}{5/4}\right)^{3/2}=\left(\dfrac2{\sqrt5}\right)^3B$ .

(C)

Topics:Biot-Savart law

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JEE Main 2022 · 24 Jun · Shift 1, Physics question 49