Question:
Two infinitely long wires are placed at \((1\text{ cm},1\text{ cm})\) and \((+1\text{ cm},-1\text{ cm})\) with \(1\text{ A}\) current in each and in the same directions perpendicular to the \(x-y\) plane. Let the magnetic field due to these current carrying wires at the origin be \(\vec{B}\). If \(B_0\) is the magnitude of the field if only one of them was present, then \(\dfrac{|\vec{B}|}{B_0}\) is:
Two infinitely long wires are placed at \((1\text{ cm},1\text{ cm})\) and \((+1\text{ cm},-1\text{ cm})\) with \(1\text{ A}\) current in each and in the same directions perpendicular to the \(x-y\) plane. Let the magnetic field due to these current carrying wires at the origin be \(\vec{B}\). If \(B_0\) is the magnitude of the field if only one of them was present, then \(\dfrac{|\vec{B}|}{B_0}\) is:
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For magnetic fields due to multiple long straight current carrying wires, first find the direction of each field using the right-hand thumb rule, then add the fields vectorially.
Updated On: Jun 26, 2026
- \(\sqrt{2}\)
- \(1\)
- \(\dfrac{1}{\sqrt{2}}\)
- \(\dfrac{1}{2\sqrt{2}}\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the positions of the wires.
The two wires are placed at \[ (1,1) \] and \[ (1,-1) \] The point where magnetic field is required is the origin: \[ (0,0) \] Distance of each wire from the origin is \[ r=\sqrt{(1)^2+(1)^2} \] \[ r=\sqrt{2}\ \text{cm} \] Thus, both wires are at equal distance from the origin.
Step 2: Magnetic field due to each wire.
For a long straight current carrying wire, the magnetic field is \[ B=\frac{\mu_0 I}{2\pi r} \] Since both wires carry equal current and are at the same distance from the origin, the magnitude of magnetic field due to each wire is the same.
Hence, \[ B_1=B_2=B_0 \]
Step 3: Find the directions of magnetic fields.
The currents are in the same direction perpendicular to the \(x-y\) plane.
Using the right-hand thumb rule, the magnetic field at the origin due to the wire at \((1,1)\) is directed along one tangent direction.
Similarly, the magnetic field at the origin due to the wire at \((1,-1)\) is directed along another tangent direction.
Because the wires are symmetrically placed about the \(x\)-axis, the two magnetic field vectors at the origin are perpendicular to each other.
Therefore, the angle between the two magnetic field vectors is \[ 90^\circ \]
Step 4: Calculate the resultant magnetic field.
Since \[ B_1=B_2=B_0 \] and the angle between them is \[ 90^\circ \] The resultant magnetic field is \[ |\vec{B}|=\sqrt{B_0^2+B_0^2} \] \[ |\vec{B}|=\sqrt{2B_0^2} \] \[ |\vec{B}|=\sqrt{2}B_0 \] Thus, \[ \frac{|\vec{B}|}{B_0}=\sqrt{2} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\sqrt{2}} \]
The two wires are placed at \[ (1,1) \] and \[ (1,-1) \] The point where magnetic field is required is the origin: \[ (0,0) \] Distance of each wire from the origin is \[ r=\sqrt{(1)^2+(1)^2} \] \[ r=\sqrt{2}\ \text{cm} \] Thus, both wires are at equal distance from the origin.
Step 2: Magnetic field due to each wire.
For a long straight current carrying wire, the magnetic field is \[ B=\frac{\mu_0 I}{2\pi r} \] Since both wires carry equal current and are at the same distance from the origin, the magnitude of magnetic field due to each wire is the same.
Hence, \[ B_1=B_2=B_0 \]
Step 3: Find the directions of magnetic fields.
The currents are in the same direction perpendicular to the \(x-y\) plane.
Using the right-hand thumb rule, the magnetic field at the origin due to the wire at \((1,1)\) is directed along one tangent direction.
Similarly, the magnetic field at the origin due to the wire at \((1,-1)\) is directed along another tangent direction.
Because the wires are symmetrically placed about the \(x\)-axis, the two magnetic field vectors at the origin are perpendicular to each other.
Therefore, the angle between the two magnetic field vectors is \[ 90^\circ \]
Step 4: Calculate the resultant magnetic field.
Since \[ B_1=B_2=B_0 \] and the angle between them is \[ 90^\circ \] The resultant magnetic field is \[ |\vec{B}|=\sqrt{B_0^2+B_0^2} \] \[ |\vec{B}|=\sqrt{2B_0^2} \] \[ |\vec{B}|=\sqrt{2}B_0 \] Thus, \[ \frac{|\vec{B}|}{B_0}=\sqrt{2} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\sqrt{2}} \]
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