The magnetic flux linked with a coil of N turns each having area of cross-section A, held in a magnetic field B with its plane parallel to the field, will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Zero
The magnetic flux through one turn is \( \phi = BA\cos\theta \), where \( \theta \) is the angle between the field \( \vec{B} \) and the normal to the plane of the coil. For N turns the flux linked is N times this.
The plane of the coil is parallel to the field, so the normal is perpendicular to the field.
\( \theta = 90^\circ \)
\( \phi_{\mathrm{linked}} = NBA\cos 90^\circ = 0 \)