Question:
Two long parallel straight metal wires \(A\) and \(B\) carrying currents \(12\text{ A}\) and \(36\text{ A}\) respectively, in the same direction are separated by \(50\text{ cm}\). The point relative to \(A\), where the resultant magnetic induction between the two wires due to the currents is zero, will be
Two long parallel straight metal wires \(A\) and \(B\) carrying currents \(12\text{ A}\) and \(36\text{ A}\) respectively, in the same direction are separated by \(50\text{ cm}\). The point relative to \(A\), where the resultant magnetic induction between the two wires due to the currents is zero, will be
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For two parallel wires carrying currents in the same direction, the neutral point lies between the wires and closer to the wire carrying smaller current.
Updated On: Jun 25, 2026
- \(90\text{ cm}\)
- \(7.5\text{ cm}\)
- \(28\text{ cm}\)
- \(12.5\text{ cm}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understand the magnetic field condition.
For a long straight current-carrying wire, magnetic field at distance \(r\) is \[ B=\frac{\mu_0 I}{2\pi r} \] Since both currents are in the same direction, the magnetic fields between the two wires are in opposite directions.
Therefore, the resultant magnetic field can be zero at a point between the two wires.
Step 2: Assume the neutral point.
Let the neutral point be at a distance \(x\) from wire \(A\).
Distance between the wires is \[ 50\text{ cm} \] So, distance of the point from wire \(B\) is \[ 50-x \]
Step 3: Equate magnetic fields due to both wires.
For resultant magnetic induction to be zero: \[ B_A=B_B \] \[ \frac{\mu_0 I_A}{2\pi x} = \frac{\mu_0 I_B}{2\pi(50-x)} \] Cancel common terms: \[ \frac{I_A}{x}=\frac{I_B}{50-x} \] Substitute: \[ I_A=12\text{ A},\quad I_B=36\text{ A} \] \[ \frac{12}{x}=\frac{36}{50-x} \]
Step 4: Solve for \(x\).
Cross-multiplying: \[ 12(50-x)=36x \] \[ 600-12x=36x \] \[ 600=48x \] \[ x=12.5\text{ cm} \]
Step 5: Final conclusion.
Therefore, the point is at a distance \[ \boxed{12.5\text{ cm}} \] from wire \(A\).
For a long straight current-carrying wire, magnetic field at distance \(r\) is \[ B=\frac{\mu_0 I}{2\pi r} \] Since both currents are in the same direction, the magnetic fields between the two wires are in opposite directions.
Therefore, the resultant magnetic field can be zero at a point between the two wires.
Step 2: Assume the neutral point.
Let the neutral point be at a distance \(x\) from wire \(A\).
Distance between the wires is \[ 50\text{ cm} \] So, distance of the point from wire \(B\) is \[ 50-x \]
Step 3: Equate magnetic fields due to both wires.
For resultant magnetic induction to be zero: \[ B_A=B_B \] \[ \frac{\mu_0 I_A}{2\pi x} = \frac{\mu_0 I_B}{2\pi(50-x)} \] Cancel common terms: \[ \frac{I_A}{x}=\frac{I_B}{50-x} \] Substitute: \[ I_A=12\text{ A},\quad I_B=36\text{ A} \] \[ \frac{12}{x}=\frac{36}{50-x} \]
Step 4: Solve for \(x\).
Cross-multiplying: \[ 12(50-x)=36x \] \[ 600-12x=36x \] \[ 600=48x \] \[ x=12.5\text{ cm} \]
Step 5: Final conclusion.
Therefore, the point is at a distance \[ \boxed{12.5\text{ cm}} \] from wire \(A\).
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