JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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 \(r << a )\)
- A \(\frac{3}{2} \frac{ a ^{2}}{ r ^{2}}\)
- B \(\frac{2}{3} \frac{ a ^{2}}{ r ^{2}}\)
- C \(\frac{2}{3} \frac{ r ^{2}}{ a ^{2}}\)
- D \(\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}\)
Step-by-step Solution
Detailed explanation
\(B _{\text {axis }}=\frac{\mu_{0} i R ^{2}}{2\left( R ^{2}+ x ^{2}\right)^{3 / 2}}\) \(B _{\text {centre }}=\frac{\mu_{0} i }{2 R }\) \(\therefore B _{\text {centre }}=\frac{\mu_{0} i }{2 a }\)…
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