MHT CET · Physics · Magnetic Effects of Current
Two concentric coplanar circular loops of radii ' \(r_1\) ' and ' \(r_2\) ' respectively carry currents ' \(\mathrm{i}_1\) ' and ' \(\mathrm{i}_2\) ' in opposite directions (one clockwise and other anticlockwise). The magnetic induction at the center of the loops is half that due to ' \(\mathrm{i}_1\) ' alone at the center. If \(r_2=2 r_1\), the value of \(\frac{i_2}{i_1}\)
- A \(\frac{1}{4}\)
- B 1
- C 2
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Magnetic field due to current \(i_1\) is given by
\(
B_1=\frac{\mu_0 i_1}{2 r_1}
\)
Similarly, \(B_2=\frac{\mu_0 i_2}{2 r_2}\)
The resultant field at the center, \(\mathrm{B}=\mathrm{B}_1-\mathrm{B}_2\)
It is given that \(B=\frac{B_1}{2}\)
\(
\begin{aligned}
& \therefore \frac{\mathrm{B}_1}{2}=\mathrm{B}_1-\mathrm{B}_2 \\
& \therefore \mathrm{B}_2=\frac{\mathrm{B}_1}{2} \\
& \therefore \frac{\mu_0 \mathrm{i}_2}{2 \mathrm{r}_2}=\frac{1}{2}\left(\frac{\mu_0 \mathrm{i}_1}{2 \mathrm{r}_1}\right) \\
& \therefore \frac{\mathrm{i}_2}{\mathrm{i}_1}=\frac{1}{2} \cdot \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{1}{2} \cdot 2=1
\end{aligned}
\)
\(
B_1=\frac{\mu_0 i_1}{2 r_1}
\)
Similarly, \(B_2=\frac{\mu_0 i_2}{2 r_2}\)
The resultant field at the center, \(\mathrm{B}=\mathrm{B}_1-\mathrm{B}_2\)
It is given that \(B=\frac{B_1}{2}\)
\(
\begin{aligned}
& \therefore \frac{\mathrm{B}_1}{2}=\mathrm{B}_1-\mathrm{B}_2 \\
& \therefore \mathrm{B}_2=\frac{\mathrm{B}_1}{2} \\
& \therefore \frac{\mu_0 \mathrm{i}_2}{2 \mathrm{r}_2}=\frac{1}{2}\left(\frac{\mu_0 \mathrm{i}_1}{2 \mathrm{r}_1}\right) \\
& \therefore \frac{\mathrm{i}_2}{\mathrm{i}_1}=\frac{1}{2} \cdot \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{1}{2} \cdot 2=1
\end{aligned}
\)
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