Question

The radius of a coil of N turns is R. If the plane of the coil is placed parallel to a uniform magnetic field B, then the flux linked with the coil is

• A. \(\pi BNR^2\)
• B. \(2\pi BNR^2\)
• C. \( \frac{\pi BNR^2}{2} \)
• D. Zero

Hint:

Be very careful with the angle in the magnetic flux formula. \(\theta\) is the angle between the B-field and the *normal* to the area, not the plane of the area itself. - If the plane is parallel to the field, \(\theta = 90^\circ\), and flux is zero. - If the plane is perpendicular to the field, \(\theta = 0^\circ\), and flux is maximum (\(NBA\)).

Answer 1

The Correct Option is D

Magnetic flux (\(\Phi\)) through a surface is defined as \( \Phi = B \cdot A = BA\cos\theta \), where B is the magnetic field strength, A is the area of the surface, and \(\theta\) is the angle between the magnetic field vector and the normal vector to the area.
For a coil with N turns, the total magnetic flux linked is \( \Phi_{total} = N(BA\cos\theta) \).
In this problem, the plane of the coil is placed *parallel* to the uniform magnetic field B.
The area vector (or normal vector) is, by definition, perpendicular to the plane of the coil.
Therefore, the angle \(\theta\) between the area vector and the magnetic field vector is \( 90^\circ \) (or \( \pi/2 \) radians).
Now, we calculate the flux using the formula.
\( \Phi_{total} = N B A \cos(90^\circ) \).
Since \( \cos(90^\circ) = 0 \), the flux is:
\( \Phi_{total} = N B A \times 0 = 0 \).
The flux linked with the coil is zero.