Question:
A wire carrying current I has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to the X-axis while the semicircular portion of radius R lies in the YZ-plane. Magnetic field at point O is:
A wire carrying current I has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to the X-axis while the semicircular portion of radius R lies in the YZ-plane. Magnetic field at point O is:
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• Semicircle at centre: B=(mu₀ I)/(4R)
• Long straight wire: B=(mu₀ I)/(2π r)
Always apply right-hand thumb rule for direction.
Updated On: Mar 19, 2026
- \(\vec{B}=-\dfrac{\mu_0 I}{4\pi R}(\hat{i}+2\hat{k})\)
- \(\vec{B}=-\dfrac{\mu_0 I}{4\pi R}(\hat{i}+2\hat{k})\)
- \(\vec{B}=\dfrac{\mu_0 I}{4\pi R}(\hat{i}-2\hat{k})\)
- vecB=(mu₀ I)/(4π R)(hati+2hatk)
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The Correct Option is D
Solution and Explanation
Step 1: Magnetic field at the centre of a semicircular loop: Bₛemi=(mu₀ I)/(4R) Direction is along +hatk (right-hand rule).
Step 2: Each long straight wire contributes: Bwire=(mu₀ I)/(2π R) Net contribution of the two straight wires is along +hati.
Step 3: Adding vector components: vecB=(mu₀ I)/(4π R)(hati+2hatk)
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