MHT CET · Physics · Magnetic Effects of Current
A circular coil of radius 'R' is carrying a current ' \(\mathrm{I}_{1}\) ' in anticlockwise sense. A long straight wire is carrying current ' \(\mathrm{I}_{2}\) ' in the negative direction of \(\mathrm{x}\) axis. Both are placed in the same plane and the distance between centre of coil and straight wire is 'd'. The magnetic field at the centre of coil will be zero for the value of 'd' equal to
- A \(\frac{\pi}{R}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)\)
- B \(\frac{\pi}{R}\left(\frac{\mathrm{I}_{2}}{\mathrm{R}_{1}}\right)\)
- C \(\frac{\mathrm{R}}{\pi}\left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}\right)\)
- D \(\frac{\mathrm{R}}{\pi}\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{R}}{\pi}\left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}\right)\)
Step-by-step Solution
Detailed explanation
(C)
Magnetic field at the centre due to coil
\(\begin{array}{l}
\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \mathrm{R}} \\
\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \pi \mathrm{d}} \\
\mathrm{B}_{1}=\mathrm{B}_{2} \\
\frac{\mu_{0} \mathrm{I}_{1}}{2 \mathrm{R}}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \pi \mathrm{d}} \\
\mathrm{d}=\frac{\mu_{0} \mathrm{I}_{2}}{\mu_{0} \mathrm{I}_{1}} \frac{2 \mathrm{R}}{2 \pi} \\
\mathrm{d}=\frac{\mathrm{I}_{2} \mathrm{R}}{\mathrm{I}_{1} \pi}
\end{array}\)
Magnetic field at the centre due to coil
\(\begin{array}{l}
\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \mathrm{R}} \\
\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \pi \mathrm{d}} \\
\mathrm{B}_{1}=\mathrm{B}_{2} \\
\frac{\mu_{0} \mathrm{I}_{1}}{2 \mathrm{R}}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \pi \mathrm{d}} \\
\mathrm{d}=\frac{\mu_{0} \mathrm{I}_{2}}{\mu_{0} \mathrm{I}_{1}} \frac{2 \mathrm{R}}{2 \pi} \\
\mathrm{d}=\frac{\mathrm{I}_{2} \mathrm{R}}{\mathrm{I}_{1} \pi}
\end{array}\)
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