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GUJCET · Physics · Moving Charges and Magnetism
A current of 10 A flows through a horizontal straight wire \(A\) with both ends rigidly fixed. Wire B is placed directly above and parallel to A. The weight per unit length of wire \(B\) is \(40 \times 10^{-3} Nm ^{-1}\) and it carries a current of 20 A . Find the distance of the wire \(B\) from the wire \(A\) so that the wire \(B\) remains at rest. Also state the direction of current flowing through it.
- A \(\frac{1}{3} \times 10^{-3} m\), both are in same direction
- B \(\frac{1}{2} \times 10^{-3} m\), both are in opposite direction
- C \(2 \times 10^{-3} m\), both are in same direction
- D \(1 \times 10^{-3} m\), both are in opposite direction
Answer & Solution
Correct Answer
(D) \(1 \times 10^{-3} m\), both are in opposite direction
Step-by-step Solution
Detailed explanation
D
\(\frac{ F }{l}=\frac{\mu_0 I _1 I _2}{2 \pi y}\)
\(\therefore 40 \times 10^{-3}=\frac{4 \pi \times 10^{-7} \times 10 \times 20}{2 \pi \times y}\)
\(\therefore \quad y=\frac{2 \times 10^{-7} \times 10 \times 20}{40 \times 10^{-3}}\)
\(y=10^{-3} m\)
Current will flow in opposite directions.
\(\frac{ F }{l}=\frac{\mu_0 I _1 I _2}{2 \pi y}\)
\(\therefore 40 \times 10^{-3}=\frac{4 \pi \times 10^{-7} \times 10 \times 20}{2 \pi \times y}\)
\(\therefore \quad y=\frac{2 \times 10^{-7} \times 10 \times 20}{40 \times 10^{-3}}\)
\(y=10^{-3} m\)
Current will flow in opposite directions.
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