Magnetic Effects & Magnetism

JEE Main level, single correct, medium

19 of 276

Question 19 of 276

The magnetic field at the centre of a current carrying circular loop of radius RR is 16μ T16 \mu \mathrm{~T}. The magnetic field at a distance x=3Rx=\sqrt{3} R on its axis from the centre is ____\_\_\_\_ μT\mu \mathrm{T}.

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Answer: (B) 2

The magnetic field at the center of a circular loop is Bcenter=μ0I2R=16μB_{center} = \frac{\mu_0 I}{2R} = 16\mu T.
On the axis at distance xx from center: B(x)=μ0IR22(R2+x2)3/2B(x) = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}.
At x=3Rx = \sqrt{3}R: B(3R)=μ0IR22(R2+3R2)3/2=μ0IR22(4R2)3/2=μ0I16RB(\sqrt{3}R) = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 I}{16R}.
The ratio: B(3R)Bcenter=1/16R1/2R=18\frac{B(\sqrt{3}R)}{B_{center}} = \frac{1/16R}{1/2R} = \frac{1}{8}.
Therefore: B(3R)=18×16μ=2μB(\sqrt{3}R) = \frac{1}{8} \times 16\mu = 2\mu T.

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