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 .......... \(m\)
Pick an option to check your answer.
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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