Question:
A circular loop of area \(0.04 \, m^2\) is rotating in a uniform magnetic field of \(0.4 \, T\). The loop rotates about its diameter which is perpendicular to the magnetic field. The magnetic flux through the loop when the plane of the coil is normal to the field and when it is making \(30^\circ\) with the field are respectively (in Wb).
A circular loop of area \(0.04 \, m^2\) is rotating in a uniform magnetic field of \(0.4 \, T\). The loop rotates about its diameter which is perpendicular to the magnetic field. The magnetic flux through the loop when the plane of the coil is normal to the field and when it is making \(30^\circ\) with the field are respectively (in Wb).
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Always remember: magnetic flux depends on angle between field and normal, not plane.
Updated On: Jun 20, 2026
- \(1.6 \times 10^{-2}, \; 0.8 \times 10^{-2}\)
- \(1.6 \times 10^{-2}, \; 1.386 \times 10^{-2}\)
- \(0, \; 0.8 \times 10^{-2}\)
- \(1.386 \times 10^{-2}, \; 0\)
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The Correct Option is B
Solution and Explanation
Step 1: Magnetic flux formula.
Magnetic flux through a loop is given by: \[ \Phi = BA \cos\theta \] where \(\theta\) is the angle between magnetic field \(B\) and the normal to the plane of the loop.
Step 2: Given values.
\[ A = 0.04 \, m^2, \quad B = 0.4 \, T \] So, \[ BA = 0.4 \times 0.04 = 0.016 = 1.6 \times 10^{-2} \]
Step 3: Case 1 — plane normal to magnetic field.
If plane is perpendicular to field, then normal is parallel to field: \[ \theta = 0^\circ \] \[ \Phi_1 = BA \cos 0^\circ = BA \] \[ \Phi_1 = 1.6 \times 10^{-2} \, Wb \]
Step 4: Case 2 — plane makes \(30^\circ\) with field.
If plane makes \(30^\circ\) with field, then normal makes: \[ \theta = 30^\circ \]
Step 5: Calculate flux at \(30^\circ\).
\[ \Phi_2 = BA \cos 30^\circ \] \[ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \] \[ \Phi_2 = 0.016 \times 0.866 \] \[ \Phi_2 = 0.013856 \, Wb \] \[ \Phi_2 \approx 1.386 \times 10^{-2} \, Wb \]
Step 6: Final interpretation.
Flux decreases as angle between field and normal increases because only the perpendicular component of magnetic field contributes to flux.
\[ \boxed{1.6 \times 10^{-2}, \; 1.386 \times 10^{-2}} \]
Magnetic flux through a loop is given by: \[ \Phi = BA \cos\theta \] where \(\theta\) is the angle between magnetic field \(B\) and the normal to the plane of the loop.
Step 2: Given values.
\[ A = 0.04 \, m^2, \quad B = 0.4 \, T \] So, \[ BA = 0.4 \times 0.04 = 0.016 = 1.6 \times 10^{-2} \]
Step 3: Case 1 — plane normal to magnetic field.
If plane is perpendicular to field, then normal is parallel to field: \[ \theta = 0^\circ \] \[ \Phi_1 = BA \cos 0^\circ = BA \] \[ \Phi_1 = 1.6 \times 10^{-2} \, Wb \]
Step 4: Case 2 — plane makes \(30^\circ\) with field.
If plane makes \(30^\circ\) with field, then normal makes: \[ \theta = 30^\circ \]
Step 5: Calculate flux at \(30^\circ\).
\[ \Phi_2 = BA \cos 30^\circ \] \[ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \] \[ \Phi_2 = 0.016 \times 0.866 \] \[ \Phi_2 = 0.013856 \, Wb \] \[ \Phi_2 \approx 1.386 \times 10^{-2} \, Wb \]
Step 6: Final interpretation.
Flux decreases as angle between field and normal increases because only the perpendicular component of magnetic field contributes to flux.
\[ \boxed{1.6 \times 10^{-2}, \; 1.386 \times 10^{-2}} \]
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