MHT CET · Physics · Magnetic Effects of Current
A coil of ' \(n\) ' turns and radius ' \(R\) ' carries a current 'I'. It is unwound and rewound to make a new coil of radius \(\frac{R}{3}\) and the same current is passed through it. The ratio of the magnetic moment of the new coil to that of the original coil is
- A 3
- B 2
- C \(\frac{1}{3}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
Let the initial magnetic moment be \(\mu_1=n_1 I A_1\)
\(=\mathrm{n}_1 \mathrm{I} \pi \mathrm{R}_1{ }^2\)
And, the final magnetic moment \(\mu_2=\mathrm{n}_2 \mathrm{I} \pi \mathrm{R}_2^2\)
Given, \(\mathrm{R}_1=\mathrm{R}, \mathrm{R}_2=\frac{\mathrm{R}}{3}\), \(\Rightarrow \mathrm{n}_2=3 \mathrm{n}_1\)
\(\therefore \quad \frac{\mu_2}{\mu_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1}\left(\frac{\mathrm{R}_2}{\mathrm{R}_1}\right)^2=3 \times\left(\frac{1}{3}\right)^2=\frac{1}{3}\)
\(=\mathrm{n}_1 \mathrm{I} \pi \mathrm{R}_1{ }^2\)
And, the final magnetic moment \(\mu_2=\mathrm{n}_2 \mathrm{I} \pi \mathrm{R}_2^2\)
Given, \(\mathrm{R}_1=\mathrm{R}, \mathrm{R}_2=\frac{\mathrm{R}}{3}\), \(\Rightarrow \mathrm{n}_2=3 \mathrm{n}_1\)
\(\therefore \quad \frac{\mu_2}{\mu_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1}\left(\frac{\mathrm{R}_2}{\mathrm{R}_1}\right)^2=3 \times\left(\frac{1}{3}\right)^2=\frac{1}{3}\)
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