Question:
Two parallel wires separated by a distance of 10 cm carry currents of 10 A and 40 A along the same direction. Where should a third current be placed to experience no magnetic force?
Two parallel wires separated by a distance of 10 cm carry currents of 10 A and 40 A along the same direction. Where should a third current be placed to experience no magnetic force?
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Set the two fields equal between the wires: \(10/r = 40/(10-r)\); the null lies closer to the weaker current.
Updated On: Jul 2, 2026
- 2 cm from the 10 A current
- 8 cm from 10 A current
- 6 cm from 10 A current
- 5 cm from 10 A current
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The Correct Option is A
Solution and Explanation
Step 1: The third wire feels no force where the net magnetic field of the other two vanishes. Because both currents flow the same way, their fields oppose only in the region between the wires, so the null point lies between them.
Step 2: Let the point be at distance \(r\) from the 10 A wire, so it is \((10-r)\) cm from the 40 A wire. Equate the field magnitudes:
\[\frac{\mu_0 (10)}{2\pi r} = \frac{\mu_0 (40)}{2\pi (10-r)}\]
Step 3: Cancel common factors and solve:
\[\frac{10}{r} = \frac{40}{10-r} \;\Rightarrow\; 10(10-r) = 40r \;\Rightarrow\; 100 = 50r\]
\[\boxed{r = 2\ \text{cm from the 10 A wire}}\]
Step 2: Let the point be at distance \(r\) from the 10 A wire, so it is \((10-r)\) cm from the 40 A wire. Equate the field magnitudes:
\[\frac{\mu_0 (10)}{2\pi r} = \frac{\mu_0 (40)}{2\pi (10-r)}\]
Step 3: Cancel common factors and solve:
\[\frac{10}{r} = \frac{40}{10-r} \;\Rightarrow\; 10(10-r) = 40r \;\Rightarrow\; 100 = 50r\]
\[\boxed{r = 2\ \text{cm from the 10 A wire}}\]
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