MHT CET · Physics · Magnetic Effects of Current
Two concentric circular coils of 'n' turns each are situated in the same plane. Their
radii are ' \(a_{1}\) ' and ' \(a_{2}{ }^{\prime}\left(a_{2}>a_{1}\right)\) and they carry currents 'I \(_{1}\) ' and ' \(\mathrm{I}_{2}\) ' respectively
\(\left(\mathrm{I}_{1}>\mathrm{I}_{2}\right)\) in opposite direction. The magnetic field at the centre is
- A \(\frac{\mu_{0} n}{2}\left[\frac{\mathrm{I}_{1} \mathrm{a}_{2}-\mathrm{I}_{2} \mathrm{a}_{1}}{\mathrm{a}_{1} \mathrm{a}_{2}}\right]\)
- B \(\frac{\mu_{0} n}{2 \mathrm{a}_{1} \mathrm{a}_{2}}\left[\mathrm{I}_{1}-\mathrm{I}_{2}\right]\)
- C \(\frac{\mu_{0} n}{2 \mathrm{I}_{1} \mathrm{I}_{2}}\left[\mathrm{a}_{2}-\mathrm{a}_{1}\right]\)
- D \(\frac{\mu_{0} n}{2}\left[\frac{\mathrm{I}_{1} \mathrm{a}_{1}-\mathrm{I}_{2} \mathrm{a}_{2}}{\mathrm{a}_{1} \mathrm{a}_{2}}\right]\)
Answer & Solution
Correct Answer
(A) \(\frac{\mu_{0} n}{2}\left[\frac{\mathrm{I}_{1} \mathrm{a}_{2}-\mathrm{I}_{2} \mathrm{a}_{1}}{\mathrm{a}_{1} \mathrm{a}_{2}}\right]\)
Step-by-step Solution
Detailed explanation
(D)
Magnetic fields due to the two coils are in opposite directions and are given by
\(\begin{array}{l}
\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{nI}_{1}}{2 \mathrm{a}_{1}} \quad \text { and } \quad \mathrm{B}_{2}=\frac{\mu_{0} \mathrm{nI}_{2}}{2 \mathrm{a}_{2}}
\end{array}\)
\(\begin{aligned} \therefore \mathrm{B} &=\mathrm{B}_{1}-\mathrm{B}_{1} \\ &=\frac{\mu_{0} \mathrm{nI}}{2}\left[\frac{\mathrm{I}_{1}}{\mathrm{a}_{1}}-\frac{\mathrm{I}_{2}}{\mathrm{a}_{2}}\right] \\ &=\frac{\mu_{0} \mathrm{nI}}{2}\left[\frac{\mathrm{I}_{1} \mathrm{a}_{2}-\mathrm{I}_{2} \mathrm{a}_{1}}{\mathrm{a}_{1} \mathrm{a}_{2}}\right] \end{aligned}\)
Magnetic fields due to the two coils are in opposite directions and are given by
\(\begin{array}{l}
\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{nI}_{1}}{2 \mathrm{a}_{1}} \quad \text { and } \quad \mathrm{B}_{2}=\frac{\mu_{0} \mathrm{nI}_{2}}{2 \mathrm{a}_{2}}
\end{array}\)
\(\begin{aligned} \therefore \mathrm{B} &=\mathrm{B}_{1}-\mathrm{B}_{1} \\ &=\frac{\mu_{0} \mathrm{nI}}{2}\left[\frac{\mathrm{I}_{1}}{\mathrm{a}_{1}}-\frac{\mathrm{I}_{2}}{\mathrm{a}_{2}}\right] \\ &=\frac{\mu_{0} \mathrm{nI}}{2}\left[\frac{\mathrm{I}_{1} \mathrm{a}_{2}-\mathrm{I}_{2} \mathrm{a}_{1}}{\mathrm{a}_{1} \mathrm{a}_{2}}\right] \end{aligned}\)
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