The magnetic field at the centre of a circular coil of radius ' $R$ ', carrying current $2 \mathrm{~A}$ is '…

The magnetic field at the centre of a circular coil of radius ' $R$ ', carrying current $2 \mathrm{~A}$ is ' $\mathrm{B}_1$ '. The magnetic field at the centre of another coil of radius ' $3 R$ ' carrying current $4 A$ is ' $B_2$ '. The ratio $B_1: B_2$ is
  1. $1: 2$
  2. $2: 1$
  3. $2: 3$
  4. $3: 2$

Solution

$\begin{aligned} \mathrm{B}_1 & =\frac{\mu_0}{4 \pi} \times \frac{2 \pi \times 2}{\mathrm{R}}=\frac{\mu_0}{\mathrm{R}} \\ \mathrm{B}_2 & =\frac{\mu_0}{4 \pi} \times \frac{2 \pi \times 4}{3 \mathrm{R}}=\frac{2 \mu_0}{3 \mathrm{R}} \\ \therefore \quad \frac{\mathrm{B}_1}{\mathrm{~B}_2} & =\frac{\left(\frac{\mu_0}{\mathrm{R}}\right)}{\frac{2 \mu_0}{(3 \mathrm{R})}}=\frac{3}{2}\end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 2)

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