KCET · Physics · Magnetic Effects of Current
Two parallel wires \( 1 \mathrm{~m} \) apart carry currents of \( 1 \mathrm{~A} \) and \( 3 \mathrm{~A} \) respectively in opposite directions. The force per unit length acting between these two wires is
- A \( 6 \times 10^{-7} \mathrm{Nm}^{-1} \) repulsive
- B \( 6 \times 10^{-7} \mathrm{Nm}^{-1} \) attractive
- C \( 6 \times 10^{-5} \mathrm{Nm}^{-1} \) repulsive
- D \( 6 \times 10^{-5} \mathrm{Nm}^{-1} \) attractive
Answer & Solution
Correct Answer
(A) \( 6 \times 10^{-7} \mathrm{Nm}^{-1} \) repulsive
Step-by-step Solution
Detailed explanation
Consider the two parallel wires as shown in the following figure.
Now, force experienced by the two wires is
\(F=\frac{\mu_{0}}{4_{\Pi}} \frac{2 I_{1} I_{2}}{d}\)
We know \( I_{1}=1 A ; I_{2}=3 A ; d=1 m \) and \( \frac{\mu_{0}}{4 \Pi}=10^{-7} \)
Therefore, \( F=\frac{10^{-7} \times 2 \times 1 \times 3}{1}=6 \times 10^{-7} \)
and since the currents are in opposite directions, therefore, force experienced will be repulsive
So, the force per unit length acting between two wires is \( 6 \times 10^{-7} \mathrm{~N} \mathrm{~m}^{-1} \) and it is repulsive.
Now, force experienced by the two wires is
\(F=\frac{\mu_{0}}{4_{\Pi}} \frac{2 I_{1} I_{2}}{d}\)
We know \( I_{1}=1 A ; I_{2}=3 A ; d=1 m \) and \( \frac{\mu_{0}}{4 \Pi}=10^{-7} \)
Therefore, \( F=\frac{10^{-7} \times 2 \times 1 \times 3}{1}=6 \times 10^{-7} \)
and since the currents are in opposite directions, therefore, force experienced will be repulsive
So, the force per unit length acting between two wires is \( 6 \times 10^{-7} \mathrm{~N} \mathrm{~m}^{-1} \) and it is repulsive.
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