MHT CET · Physics · Magnetic Effects of Current
A current carrying circular loop of radius ' \(R\) ' and current carrying long straight wire are placed in the same plane. The current through circular loop and long straight wire are ' \(I_C\) ' and ' 1 w ' respectively. The perpendicular distance between centre of the circular loop and wire is ' d '. The magnetic field at the centre of the loop will be zero when separation ' \(d\) ' is equal to
- A \(\frac{\mathrm{RI}_{\mathrm{w}}}{\pi \mathrm{I}_{\mathrm{C}}}\)
- B \(\frac{R I_C}{\pi I_w}\)
- C \(\frac{\pi \mathrm{I}_{\mathrm{C}}}{\mathrm{RI}_{\mathrm{w}}}\)
- D \(\frac{\pi \mathrm{I}_{\mathrm{w}}}{\mathrm{RI}_{\mathrm{C}}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{RI}_{\mathrm{w}}}{\pi \mathrm{I}_{\mathrm{C}}}\)
Step-by-step Solution
Detailed explanation
Magnetic field due to straight wire \(=\frac{\mu_0 I_w}{2 \pi d}\)
Magnetic field due to circular loop \(=\frac{\mu_0 I_C}{2 R}\)
As magnetic field at centre is zero
\(\begin{array}{ll}
& \frac{\mu_0 I_w}{2 \pi d}=\frac{\mu_0 I_c}{2 R} \\
\therefore \quad & d=\frac{R I_w}{\pi I_c}
\end{array}\)
Magnetic field due to circular loop \(=\frac{\mu_0 I_C}{2 R}\)
As magnetic field at centre is zero
\(\begin{array}{ll}
& \frac{\mu_0 I_w}{2 \pi d}=\frac{\mu_0 I_c}{2 R} \\
\therefore \quad & d=\frac{R I_w}{\pi I_c}
\end{array}\)
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