Question:
A long wire lies along \(x\)-axis and carries a current of \(40\,\text{A}\) in positive \(x\)-direction. A second long wire is perpendicular to the \(xy\)-plane, passes through point \((3.0\,\text{m})\hat{j}\) and carries a current along positive \(z\)-direction. If the magnitude of resultant magnetic field at the point \((2.0\,\text{m})\hat{j}\) is \(5\times 10^{-6}\,\text{T}\), then the current in the second wire is \((\mu_0=4\pi\times 10^{-7}\,\text{SI unit})\)
A long wire lies along \(x\)-axis and carries a current of \(40\,\text{A}\) in positive \(x\)-direction. A second long wire is perpendicular to the \(xy\)-plane, passes through point \((3.0\,\text{m})\hat{j}\) and carries a current along positive \(z\)-direction. If the magnitude of resultant magnetic field at the point \((2.0\,\text{m})\hat{j}\) is \(5\times 10^{-6}\,\text{T}\), then the current in the second wire is \((\mu_0=4\pi\times 10^{-7}\,\text{SI unit})\)
Show Hint
For a long straight current-carrying wire,
\[
B=\frac{\mu_0 I}{2\pi r}.
\]
If two magnetic fields are perpendicular, use
\[
B_{\text{net}}=\sqrt{B_1^2+B_2^2}.
\]
Updated On: Jun 18, 2026
- \(30\,\text{A}\)
- \(15\,\text{A}\)
- \(25\,\text{A}\)
- \(7.5\,\text{A}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Magnetic field due to the first wire.
For a long straight wire, \[ B=\frac{\mu_0 I}{2\pi r} \] The first wire lies along the \(x\)-axis.
The point is at \[ (2.0\,\text{m})\hat{j}, \] so its perpendicular distance from the \(x\)-axis is \[ r_1=2\,\text{m}. \] Thus, \[ B_1=\frac{(4\pi\times 10^{-7})(40)}{2\pi(2)} \] \[ B_1=4\times 10^{-6}\,\text{T}. \]
Step 2: Magnetic field due to the second wire.
The second wire passes through \[ (3.0\,\text{m})\hat{j} \] and the field is required at \[ (2.0\,\text{m})\hat{j}. \] Hence, the perpendicular distance from the second wire is \[ r_2=3-2=1\,\text{m}. \] Let the current in the second wire be \(I\). Then, \[ B_2=\frac{\mu_0 I}{2\pi r_2} \] \[ B_2=\frac{(4\pi\times 10^{-7})I}{2\pi(1)} \] \[ B_2=2\times 10^{-7}I. \]
Step 3: Use resultant magnetic field.
The magnetic fields due to the two wires are mutually perpendicular.
Therefore, \[ B_{\text{net}}^2=B_1^2+B_2^2 \] Given, \[ B_{\text{net}}=5\times 10^{-6}\,\text{T}. \] So, \[ (5\times 10^{-6})^2=(4\times 10^{-6})^2+B_2^2 \] \[ 25\times 10^{-12}=16\times 10^{-12}+B_2^2 \] \[ B_2^2=9\times 10^{-12} \] \[ B_2=3\times 10^{-6}\,\text{T}. \]
Step 4: Find the current in the second wire.
Using \[ B_2=2\times 10^{-7}I, \] we get \[ 3\times 10^{-6}=2\times 10^{-7}I \] \[ I=\frac{3\times 10^{-6}}{2\times 10^{-7}} \] \[ I=15\,\text{A}. \]
Step 5: Final conclusion.
Therefore, the current in the second wire is \[ \boxed{15\,\text{A}} \]
For a long straight wire, \[ B=\frac{\mu_0 I}{2\pi r} \] The first wire lies along the \(x\)-axis.
The point is at \[ (2.0\,\text{m})\hat{j}, \] so its perpendicular distance from the \(x\)-axis is \[ r_1=2\,\text{m}. \] Thus, \[ B_1=\frac{(4\pi\times 10^{-7})(40)}{2\pi(2)} \] \[ B_1=4\times 10^{-6}\,\text{T}. \]
Step 2: Magnetic field due to the second wire.
The second wire passes through \[ (3.0\,\text{m})\hat{j} \] and the field is required at \[ (2.0\,\text{m})\hat{j}. \] Hence, the perpendicular distance from the second wire is \[ r_2=3-2=1\,\text{m}. \] Let the current in the second wire be \(I\). Then, \[ B_2=\frac{\mu_0 I}{2\pi r_2} \] \[ B_2=\frac{(4\pi\times 10^{-7})I}{2\pi(1)} \] \[ B_2=2\times 10^{-7}I. \]
Step 3: Use resultant magnetic field.
The magnetic fields due to the two wires are mutually perpendicular.
Therefore, \[ B_{\text{net}}^2=B_1^2+B_2^2 \] Given, \[ B_{\text{net}}=5\times 10^{-6}\,\text{T}. \] So, \[ (5\times 10^{-6})^2=(4\times 10^{-6})^2+B_2^2 \] \[ 25\times 10^{-12}=16\times 10^{-12}+B_2^2 \] \[ B_2^2=9\times 10^{-12} \] \[ B_2=3\times 10^{-6}\,\text{T}. \]
Step 4: Find the current in the second wire.
Using \[ B_2=2\times 10^{-7}I, \] we get \[ 3\times 10^{-6}=2\times 10^{-7}I \] \[ I=\frac{3\times 10^{-6}}{2\times 10^{-7}} \] \[ I=15\,\text{A}. \]
Step 5: Final conclusion.
Therefore, the current in the second wire is \[ \boxed{15\,\text{A}} \]
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