Question

A rectangular coil of area A is kept in a uniform magnetic field \(\vec{B}\) such that the plane of the coil makes an angle \(\alpha\) with \(\vec{B}\). The magnetic flux linked with the coil is:

• A. \(BA \sin \alpha\)
• B. \(BA \cos \alpha\)
• C. \(BA\)
• D. zero

Hint:

When calculating magnetic flux, ensure you use the angle between the magnetic field \(\vec{B}\) and the normal to the plane of the coil (area vector \(\vec{A}\)). The formula is \(\phi = B A \cos \theta\), where \(\theta\) is this angle.

Answer 1

The Correct Option is B

Step 1: Recall the magnetic flux formula.
Magnetic flux \(\phi\) is calculated as: \[\phi = \vec{B} \cdot \vec{A}\]Here, \(\vec{B}\) represents the magnetic field, \(\vec{A}\) is the area vector of the coil, and the dot product considers the angle between them.Step 2: Ascertain the angle between \(\vec{B}\) and \(\vec{A}\).
The area vector \(\vec{A}\) is perpendicular to the coil's plane and has a magnitude of \(A\). The problem specifies that the coil's plane forms an angle \(\alpha\) with \(\vec{B}\). Consequently, the angle between \(\vec{B}\) and the normal to the plane (which is \(\vec{A}\)) is: \[\theta = 90^\circ - \alpha\]Step 3: Calculate the magnetic flux.
The dot product is expressed as: \[\vec{B} \cdot \vec{A} = B A \cos \theta\]Substituting \(\theta = 90^\circ - \alpha\): \[\cos (90^\circ - \alpha) = \sin \alpha\]However, a standard convention in physics problems is to define the given angle \(\alpha\) as the angle between the magnetic field \(\vec{B}\) and the normal to the coil's plane (the area vector \(\vec{A}\)). Assuming this standard interpretation, the angle \(\theta\) is directly \(\alpha\). Therefore:\[\phi = B A \cos \alpha\]Step 4: Identify the matching option.
The calculated flux is \(BA \cos \alpha\), which corresponds to option (B).