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

The fractional change in the magnetic field intensity at a distance \('r'\) from centre on the axis of current carrying coil of radius \('a'\) to the magnetic field intensity at the centre of the same coil is : (Take \(r << a )\)

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Answer & Solution

Correct Answer

(D) \(\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}\)

Step-by-step Solution

Detailed explanation

\(B _{\text {axis }}=\frac{\mu_{0} i R ^{2}}{2\left( R ^{2}+ x ^{2}\right)^{3 / 2}}\)

\(B _{\text {centre }}=\frac{\mu_{0} i }{2 R }\)

\(\therefore B _{\text {centre }}=\frac{\mu_{0} i }{2 a }\)

\(\therefore B _{ axis }=\frac{\mu_{0} ia ^{2}}{2\left( a ^{2}+ r ^{2}\right)^{3 / 2}}\)

\(\therefore\) fractional change in magnetic field \(=\)

\(\frac{\frac{\mu_{0} i }{2 a }-\frac{\mu_{0} ia ^{2}}{2\left( a ^{2}+ r ^{2}\right)^{3 / 2}}}{{\frac{\mu_{0} i }{2 a }}}=1-\frac{1}{\left[1+\left(\frac{ r ^{2}}{ a ^{2}}\right)\right]^{3 / 2}}\)

\(\approx 1-\left[1-\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}\right]=\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}\)

Note : \(\left(1+\frac{ r ^{2}}{ a ^{2}}\right)^{-3 / 2} \approx\left(1-\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}\right)\)

[True only if \(r << a ]\)

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