KCET · Physics · Magnetic Effects of Current
A conducting wire carrying current is arranged as shown. The magnetic field at ' \( O \) '
- A \( \frac{\mu_{0} I}{12}\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)
- B \( \frac{\mu_{0} I}{12}\left[\frac{1}{R_{1}}+\frac{1}{R_{2}}\right] \)
- C \( \frac{\mu_{0} I}{6}\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)
- D \( \frac{\mu_{0} I}{6}\left[\frac{1}{R_{1}}+\frac{1}{R_{2}}\right] \)
Answer & Solution
Correct Answer
(A) \( \frac{\mu_{0} I}{12}\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)
Step-by-step Solution
Detailed explanation
Magnetic field at \( O \) due to inner current carrying wire is
\(B_{1}=\frac{\mu_{0}}{4 \Pi} \frac{2 \Pi(1 / 6) I}{R_{1}}=\frac{\mu_{0}}{12} \frac{I}{R_{1}}\)
\(B_{2}=\frac{\mu_{0}}{4 \Pi} \frac{2 \Pi(1 / 6) I}{R_{2}}=\frac{\mu_{0}}{12} \frac{I}{R_{2}}\)
Therefore, net magnetic field at \( \mathrm{O} \) is given as
\(B=B_{1}-B_{2}=\frac{\mu_{0}}{12} \frac{I}{R_{1}}-\frac{\mu_{0}}{12} \frac{I}{R_{2}}=\frac{\mu_{0} I}{12}\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(B_{1}=\frac{\mu_{0}}{4 \Pi} \frac{2 \Pi(1 / 6) I}{R_{1}}=\frac{\mu_{0}}{12} \frac{I}{R_{1}}\)
\(B_{2}=\frac{\mu_{0}}{4 \Pi} \frac{2 \Pi(1 / 6) I}{R_{2}}=\frac{\mu_{0}}{12} \frac{I}{R_{2}}\)
Therefore, net magnetic field at \( \mathrm{O} \) is given as
\(B=B_{1}-B_{2}=\frac{\mu_{0}}{12} \frac{I}{R_{1}}-\frac{\mu_{0}}{12} \frac{I}{R_{2}}=\frac{\mu_{0} I}{12}\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
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