The magnetic field at the centre of a current carrying circular loop of radius R is 16. The magnetic field at a distance x = √(3) R on its axis from the centre is. (a) 8 (b) 2 (c) 4 (d) 2√(2)

Answer: (b). (Bx)/(Bc) = R³(R²+x²)3/2 x = √(3)R ⇒ (4R²)3/2 = 8R³, so Bx = Bc/8 = 16/8 So the answer is (b) 2.

The magnetic field at the centre of a current carrying circular loop… | JEE Main 2026 (28 Jan, Shift 1)

Q4JEE Main 2026 · 28 JAN · SHIFT 1Easy

Biot-Savart Law & Circular Coils

The magnetic field at the centre of a current carrying circular loop of radius $R$ is $16\ \mu\text{T}$ . The magnetic field at a distance $x = \sqrt{3} R$ on its axis from the centre is

$\underline{\qquad}\ \mu\text{T}$ .

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Answer: (b)

$\frac{B_x}{B_c} = \frac{R^3}{(R^2+x^2)^{3/2}}$
$x = \sqrt{3}R \Rightarrow (4R^2)^{3/2} = 8R^3$ , so $B_x = \frac{B_c}{8} = \frac{16}{8}$
So the answer is (b) $2\ \mu\text{T}$ .

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