Question:

The figure given below shows three straight long parallel conductors kept in x-y plane, carrying currents 2I, I and 3I respectively as shown in figure. Find the magnitude and direction of net magnetic force acting on unit length of conductor 1, due to conductors 2 and 3.

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Alternatively, use Lorentz force per unit length $\vec{f} = I_1 (\hat{i} \times \vec{B}_{net})$.
$\vec{f} = 2I \hat{i} \times \left(-\frac{\mu_0 I}{4\pi d} \hat{k}\right) = +\frac{\mu_0 I^2}{2\pi d} \hat{j}$. Both methods yield identical results!
Updated On: Sep 14, 2026
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Solution and Explanation

Concept:
• Force per unit length between two parallel long straight currents $I_1$ and $I_2$ separated by distance $r$ is $f = \frac{\mu_0 I_1 I_2}{2\pi r}$.

• Parallel currents in the same direction attract each other; antiparallel currents in opposite directions repel each other.

Step 1:
Force per unit length due to Conductor 2
Current in conductor 1 is $2I$ along $+x$. Current in conductor 2 is $I$ along $+x$.
Since currents flow in the same direction, force is attractive (towards conductor 2, i.e., downwards along $-\hat{j}$):
\[ \vec{f}_{12} = \frac{\mu_0 (2I)(I)}{2\pi d} (-\hat{j}) = -\frac{\mu_0 I^2}{\pi d} \hat{j} \]

Step 2:
Force per unit length due to Conductor 3
Current in conductor 1 is $2I$ along $+x$. Current in conductor 3 is $3I$ along $-x$.
Since currents flow in opposite directions, force is repulsive (away from conductor 3, i.e., upwards along $+\hat{j}$):
\[ \vec{f}_{13} = \frac{\mu_0 (2I)(3I)}{2\pi (2d)} (+\hat{j}) = +\frac{6 \mu_0 I^2}{4\pi d} \hat{j} = +\frac{3 \mu_0 I^2}{2\pi d} \hat{j} \]

Step 3:
Calculate net force per unit length on conductor 1
\[ \vec{f}_{net} = \vec{f}_{12} + \vec{f}_{13} \]
\[ \vec{f}_{net} = \left( \frac{3 \mu_0 I^2}{2\pi d} - \frac{\mu_0 I^2}{\pi d} \right) \hat{j} \]
Take common denominator $2\pi d$:
\[ \vec{f}_{net} = \left( \frac{3 \mu_0 I^2 - 2 \mu_0 I^2}{2\pi d} \right) \hat{j} = +\frac{\mu_0 I^2}{2\pi d} \hat{j} \]

Step 4:
Conclusion
Magnitude of net magnetic force per unit length on conductor 1 is $f_{net} = \frac{\mu_0 I^2}{2\pi d}$, directed upwards along the positive y-axis ($+\hat{j}$).
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