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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2021
  • Hard
  • MCQ

Magnetic fields at two points on the axis of a circular coil at a distance of \(0.05\, m\) and \(0.2\, m\) from the centre are in the ratio \(8: 1.\) The radius of coil is .......... \(m\)

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Answer & Solution

Correct Answer

(B) \(0.1\)

Step-by-step Solution

Detailed explanation

We know, the magnetic field on the axis of a current carrying circular ring is given by

\(B =\frac{\mu_{0}}{4 \pi} \frac{2 NIA }{\left( R ^{2}+ x ^{2}\right)^{3 / 2}}\)

\(\therefore \frac{ B _{1}}{ B _{2}}=\frac{8}{1}=\left[\frac{ R ^{2}+(0.2)^{2}}{ R ^{2}+(0.05)^{2}}\right]^{3 / 2}\)

\(4\left[ R ^{2}+(0.05)^{2}\right]=\left[ R ^{2}+(0.2)^{2}\right]\)

\(4 R ^{2}- R ^{2}=(0.2)^{2}-4 \times(0.05)^{2}\)

\(4 R ^{2}- R ^{2}=(0.2)^{2}-(0.1)^{2}\)

\(3 R ^{2}=0.3 \times 0.1\)

\(R ^{2}=(0.1)^{2} \Rightarrow R =0.1\)

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