KCET · Physics · Magnetic Effects of Current
\(\mathrm{PQ}\) and RS are long parallel conductors separated by certain distance. \(M\) is the midpoint between them (see the figure). The net magnetic field at \(M\) is B. Now, the current 2 A is switched off. The field at \(M\) now becomes
- A \(2 \mathrm{~B}\)
- B \(\mathrm{B}\)
- C \(\frac{\mathrm{B}}{2}\)
- D \(3 B\)
Answer & Solution
Correct Answer
(B) \(\mathrm{B}\)
Step-by-step Solution
Detailed explanation
Magnetic field at midpoint \(M\) in first case is
\(\mathrm{B}=\mathrm{B}_{\mathrm{PQ}}-\mathrm{B}_{\mathrm{RS}}\)
\(\left(\because \mathrm{B}_{\mathrm{PQ}}\right.\) and \(\mathrm{B}_{\mathrm{RS}}\) are in opposite directions)
\(=\frac{4 \mu_{0}}{4 \pi \mathrm{d}}-\frac{2 \mu_{0}}{4 \pi \mathrm{d}}\)
\(=\frac{2 \mu_{0}}{4 \pi \mathrm{d}}\)
When the current \(2 \mathrm{~A}\) is switched off, the net magnetic field at \(\mathrm{M}\) is due to current \(1 \mathrm{~A}\)
\(\mathrm{B}^{\prime}=\frac{\mu_{0} \times 2 \times 1}{4 \pi \mathrm{d}}=\mathrm{B}\)
\(\mathrm{B}=\mathrm{B}_{\mathrm{PQ}}-\mathrm{B}_{\mathrm{RS}}\)
\(\left(\because \mathrm{B}_{\mathrm{PQ}}\right.\) and \(\mathrm{B}_{\mathrm{RS}}\) are in opposite directions)
\(=\frac{4 \mu_{0}}{4 \pi \mathrm{d}}-\frac{2 \mu_{0}}{4 \pi \mathrm{d}}\)
\(=\frac{2 \mu_{0}}{4 \pi \mathrm{d}}\)
When the current \(2 \mathrm{~A}\) is switched off, the net magnetic field at \(\mathrm{M}\) is due to current \(1 \mathrm{~A}\)
\(\mathrm{B}^{\prime}=\frac{\mu_{0} \times 2 \times 1}{4 \pi \mathrm{d}}=\mathrm{B}\)
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