JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
The magnetic field at the centre of a circular coil of radius \(I\), due to current I flowing through it, is \(B\). The magnetic field at a point along the axis at a distance \(\frac{r}{2}\) from the centre is
- A \(B / 2\)
- B \(2 B\)
- C \(\left(\frac{2}{\sqrt{5}}\right)^{3} B\)
- D \(\left(\frac{2}{\sqrt{3}}\right)^{3} B\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{2}{\sqrt{5}}\right)^{3} B\)
Step-by-step Solution
Detailed explanation
\(B_{C}=\frac{\mu_{0} I}{2 r}, B_{a}=\frac{\mu_{0} I^{2}}{2\left(x^{2}+r^{2}\right)^{3 / 2}}\) \(At \,\,x =\frac{r}{2}\) \(B _{ a }=\frac{\mu_{0} Ir ^{2}}{2\left(\frac{ r ^{2}}{4}+ r ^{2}\right)^{3 / 2}}\)…
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