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Question
A straight, how wire carries a current of 20 A. Another wire carrying equal current is placed parallel to it. If the force acting on a length of 10 cm of the second wire is 2.0 × 10−5 N, what is the separation between them?
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Solution
Given:
Magnitude of current in both wires, i1 = i2 = 20 A
Force acting on 0.1 m of the second wire, F = 2.0 × 10−5 N
∴ Force per unit length = \[\frac{2 \times {10}^{- 5}}{0 . 1}\] = 2.0 × 10−4 N/m
Now,
Let the separation between the two wires be d.
Thus, the force per unit length is given by
\[\frac{F}{l} = \frac{\mu_0 i_1 i_2}{2\pi d}\]
\[ \Rightarrow 2 . 0 \times {10}^{- 4} = \frac{2 \times {10}^{- 7} \times 20 \times 20}{d}\]
\[ \Rightarrow d = 0 . 4 \] m = 40 cm
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