MHT CET · Physics · Magnetic Effects of Current
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
- A \(1: 2\)
- B \(2: 1\)
- C \(2: 3\)
- D \(3: 2\)
Answer & Solution
Correct Answer
(D) \(3: 2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{B}_1 & =\frac{\mu_0}{4 \pi} \times \frac{2 \pi \times 2}{\mathrm{R}}=\frac{\mu_0}{\mathrm{R}} \\ \mathrm{B}_2 & =\frac{\mu_0}{4 \pi} \times \frac{2 \pi \times 4}{3 \mathrm{R}}=\frac{2 \mu_0}{3 \mathrm{R}} \\ \therefore \quad \frac{\mathrm{B}_1}{\mathrm{~B}_2} & =\frac{\left(\frac{\mu_0}{\mathrm{R}}\right)}{\frac{2 \mu_0}{(3 \mathrm{R})}}=\frac{3}{2}\end{aligned}\)
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