KCET · Physics · Magnetic Effects of Current
The magnetic field at the centre of a circular current carrying conductor of radius \(r\) is \(\mathrm{B}_{\mathrm{c}}\). The magnetic field on its axis at a distance \(r\) from the centre is \(\mathrm{B}_{\mathrm{a}}\). The value of \(\mathrm{B}_{\mathrm{c}}: \mathrm{B}_{\mathrm{a}}\) will be
- A \(1: \sqrt{2}\)
- B \(1: 2 \sqrt{2}\)
- C \(2 \sqrt{2}: 1\)
- D \(\sqrt{2}: 1\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
Magnetic induction at the centre of the coil of radius is
\(\mathrm{B}_{\mathrm{c}}=\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}}\)
Magnetic induction on the axial line of a circular coil at a distance from the centre is
\(B_{a}=\frac{\mu_{0} n^{2} I}{2\left(r^{2}+x^{2}\right)^{3 / 2}}\)
Given \(x=r\)
\(B_{a}=\frac{\mu_{0} n r^{2} I}{2\left(2 r^{2}\right)^{3 / 2}}\)
From Eqs. (i) and (ii), we get
\(\frac{\mathrm{B}_{\mathrm{c}}}{\mathrm{B}_{\mathrm{a}}}=\frac{2 \sqrt{2}}{1}\)
\(\mathrm{B}_{\mathrm{c}}=\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}}\)
Magnetic induction on the axial line of a circular coil at a distance from the centre is
\(B_{a}=\frac{\mu_{0} n^{2} I}{2\left(r^{2}+x^{2}\right)^{3 / 2}}\)
Given \(x=r\)
\(B_{a}=\frac{\mu_{0} n r^{2} I}{2\left(2 r^{2}\right)^{3 / 2}}\)
From Eqs. (i) and (ii), we get
\(\frac{\mathrm{B}_{\mathrm{c}}}{\mathrm{B}_{\mathrm{a}}}=\frac{2 \sqrt{2}}{1}\)
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