Question:
An infinitely long wire carrying \(1\ \text{A}\) current in the \(+Z\) direction is placed at \((1\ \text{cm},1\ \text{cm})\). Another wire carrying \(1\ \text{A}\) in \(+X\) direction is placed at \(y=1\ \text{cm}\). If the magnetic field due to this configuration at the origin is \(\vec{B}\). Let \(B_0\) be the magnitude of the field if only the wire at \((1\ \text{cm},1\ \text{cm})\) was present, then \(\frac{\vec{B}}{B_0}\) is
An infinitely long wire carrying \(1\ \text{A}\) current in the \(+Z\) direction is placed at \((1\ \text{cm},1\ \text{cm})\). Another wire carrying \(1\ \text{A}\) in \(+X\) direction is placed at \(y=1\ \text{cm}\). If the magnetic field due to this configuration at the origin is \(\vec{B}\). Let \(B_0\) be the magnitude of the field if only the wire at \((1\ \text{cm},1\ \text{cm})\) was present, then \(\frac{\vec{B}}{B_0}\) is
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For an infinitely long straight current-carrying wire,
\[
B=\frac{\mu_0 I}{2\pi r}.
\]
The direction of magnetic field is found using the right-hand thumb rule.
Updated On: Jun 26, 2026
- \(\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},-\sqrt2\right)\)
- \(\left(\frac{1}{2},\frac{1}{2},-1\right)\)
- \(\left(\sqrt2,\sqrt2,-\sqrt2\right)\)
- \(\left(\frac{1}{2\sqrt2},\frac{1}{2\sqrt2},-\frac{1}{2}\right)\)
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The Correct Option is A
Solution and Explanation
Step 1: Magnetic field due to the wire along \(+Z\)-axis.
The first wire is parallel to \(Z\)-axis and passes through \[ (1\ \text{cm},1\ \text{cm}). \] Distance of the origin from this wire is \[ r_1=\sqrt{(1)^2+(1)^2}\ \text{cm} =\sqrt2\ \text{cm}. \] So, \[ B_0=\frac{\mu_0 I}{2\pi r_1} = \frac{\mu_0 I}{2\pi\sqrt2\ \text{cm}}. \] Using the right-hand rule, the direction of this field at the origin is \[ \frac{1}{\sqrt2}\hat{i}-\frac{1}{\sqrt2}\hat{j}. \] Therefore, \[ \frac{\vec{B}_1}{B_0} = \left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},0\right). \]
Step 2: Magnetic field due to the wire along \(+X\)-axis.
The second wire carries current in the \(+X\) direction and is placed at \[ y=1\ \text{cm}. \] The distance of the origin from this wire is \[ r_2=1\ \text{cm}. \] Thus, \[ B_2=\frac{\mu_0 I}{2\pi r_2} = \frac{\mu_0 I}{2\pi(1\ \text{cm})}. \] Now, \[ \frac{B_2}{B_0} = \frac{\frac{\mu_0 I}{2\pi(1\ \text{cm})}} {\frac{\mu_0 I}{2\pi\sqrt2\ \text{cm}}} = \sqrt2. \] Using the right-hand rule, the field due to this wire at the origin is along negative \(Z\)-direction.
Therefore, \[ \frac{\vec{B}_2}{B_0} = (0,0,-\sqrt2). \]
Step 3: Add the magnetic field vectors.
The total magnetic field is \[ \vec{B}=\vec{B}_1+\vec{B}_2. \] Hence, \[ \frac{\vec{B}}{B_0} = \left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},0\right) + (0,0,-\sqrt2). \] \[ \frac{\vec{B}}{B_0} = \left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},-\sqrt2\right). \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},-\sqrt2\right)} \] Hence, the correct option is \[ \boxed{(1)} \]
The first wire is parallel to \(Z\)-axis and passes through \[ (1\ \text{cm},1\ \text{cm}). \] Distance of the origin from this wire is \[ r_1=\sqrt{(1)^2+(1)^2}\ \text{cm} =\sqrt2\ \text{cm}. \] So, \[ B_0=\frac{\mu_0 I}{2\pi r_1} = \frac{\mu_0 I}{2\pi\sqrt2\ \text{cm}}. \] Using the right-hand rule, the direction of this field at the origin is \[ \frac{1}{\sqrt2}\hat{i}-\frac{1}{\sqrt2}\hat{j}. \] Therefore, \[ \frac{\vec{B}_1}{B_0} = \left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},0\right). \]
Step 2: Magnetic field due to the wire along \(+X\)-axis.
The second wire carries current in the \(+X\) direction and is placed at \[ y=1\ \text{cm}. \] The distance of the origin from this wire is \[ r_2=1\ \text{cm}. \] Thus, \[ B_2=\frac{\mu_0 I}{2\pi r_2} = \frac{\mu_0 I}{2\pi(1\ \text{cm})}. \] Now, \[ \frac{B_2}{B_0} = \frac{\frac{\mu_0 I}{2\pi(1\ \text{cm})}} {\frac{\mu_0 I}{2\pi\sqrt2\ \text{cm}}} = \sqrt2. \] Using the right-hand rule, the field due to this wire at the origin is along negative \(Z\)-direction.
Therefore, \[ \frac{\vec{B}_2}{B_0} = (0,0,-\sqrt2). \]
Step 3: Add the magnetic field vectors.
The total magnetic field is \[ \vec{B}=\vec{B}_1+\vec{B}_2. \] Hence, \[ \frac{\vec{B}}{B_0} = \left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},0\right) + (0,0,-\sqrt2). \] \[ \frac{\vec{B}}{B_0} = \left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},-\sqrt2\right). \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},-\sqrt2\right)} \] Hence, the correct option is \[ \boxed{(1)} \]
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