KCET · Physics · Magnetic Effects of Current
A circular coil of wire of radius \(r\) has \(n\) turns and carries a current \(I\). The magnetic induction \(B\) at a point on the axis of the coil at a distance \(\sqrt{3} r\) from its centre is
- A \(\frac{\mu_0 n I}{8 r}\)
- B \(\frac{\mu_0 n I}{16 r}\)
- C \(\frac{\mu_0 n l}{4 r}\)
- D \(\frac{\mu_0 n I}{32 r}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mu_0 n I}{16 r}\)
Step-by-step Solution
Detailed explanation
Magnetic field on the axis current carrying circular coil,
\(B=\frac{\mu_0 I r^2 n}{2\left(x^2+r^2\right)^{3 / 2}}\)
where, \(I=\) current in the coil,
\(n=\) number of turns,
\(r=\) radius of coil.
and \(x=\) distance of the point of observation from centre of coil.
Here, \(x=\sqrt{3} r\)
\(\therefore B =\frac{\mu_0 I r^2 n}{2\left[(\sqrt{3} r)^2+r^2\right]^{3 / 2}}=\frac{\mu_0 I r^2 n}{2\left[3 r^2+r^2\right]^{3 / 2}} \)
\( =\frac{\mu_0 I T n}{2\left(4 r^2\right)^{3 / 2}}=\frac{\mu_0 I r^2 n}{2\left[(2 r)^2\right]^{3 / 2}}=\frac{\mu_0 I r^2 n}{2 \times 8 r^3}=\frac{\mu_0 n I}{16 r}\)
\(B=\frac{\mu_0 I r^2 n}{2\left(x^2+r^2\right)^{3 / 2}}\)
where, \(I=\) current in the coil,
\(n=\) number of turns,
\(r=\) radius of coil.
and \(x=\) distance of the point of observation from centre of coil.
Here, \(x=\sqrt{3} r\)
\(\therefore B =\frac{\mu_0 I r^2 n}{2\left[(\sqrt{3} r)^2+r^2\right]^{3 / 2}}=\frac{\mu_0 I r^2 n}{2\left[3 r^2+r^2\right]^{3 / 2}} \)
\( =\frac{\mu_0 I T n}{2\left(4 r^2\right)^{3 / 2}}=\frac{\mu_0 I r^2 n}{2\left[(2 r)^2\right]^{3 / 2}}=\frac{\mu_0 I r^2 n}{2 \times 8 r^3}=\frac{\mu_0 n I}{16 r}\)
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