Magnetic Effects & Magnetism

JEE Main level, single correct, medium

90 of 276

Question 90 of 276

Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field B1\mathrm{B}_1, at its centre when a current I\mathrm{I} passes through it. The ratio B1:B2\mathrm{B}_1: \mathrm{B}_2 is:

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Answer: (B) 1:31: 3

For loop B=μ0nI2aB=\frac{\mu_0 \mathrm{nI}}{2 \mathrm{a}} where, aa is the radius of loop.
Then, B1=μ0I2aB_1=\frac{\mu_0 I}{2 a}
Now, for coil B=μ0I4π⋅2nAx3\mathrm{B}=\frac{\mu_0 \mathrm{I}}{4 \pi} \cdot \frac{2 \mathrm{nA}}{\mathrm{x}^3} at the centre x=x= radius of loop
B2=μ04π⋅2×3×(I/3)×π(a/3)2(a/3)3=μ0⋅3I2a∴B1 B2=μ0I/2aμ0⋅3I/2aB1:B2=1:3\begin{aligned} \mathrm{B}_2 &=\frac{\mu_0}{4 \pi} \cdot \frac{2 \times 3 \times(\mathrm{I} / 3) \times \pi(\mathrm{a} / 3)^2}{(\mathrm{a} / 3)^3} \\ &=\frac{\mu_0 \cdot 3 \mathrm{I}}{2 \mathrm{a}} \\ \therefore & \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mu_0 \mathrm{I} / 2 \mathrm{a}}{\mu_0 \cdot 3 \mathrm{I} / 2 \mathrm{a}} \\ \mathrm{B}_1: \mathrm{B}_2=1: 3 \end{aligned}

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