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: Understand the magnetic flux formula.
The magnetic flux \(\phi\) through a coil is given by: \[ \phi = \vec{B} \cdot \vec{A} \] where \(\vec{B}\) is the magnetic field, \(\vec{A}\) is the area vector of the coil, and the dot product accounts for the angle between them. Step 2: Determine the angle between \(\vec{B}\) and \(\vec{A}\).
The area vector \(\vec{A}\) is perpendicular to the plane of the coil, with magnitude \(A\). The problem states that the plane of the coil makes an angle \(\alpha\) with \(\vec{B}\). Therefore, the angle between \(\vec{B}\) and the normal to the plane (i.e., \(\vec{A}\)) is: \[ \theta = 90^\circ - \alpha \] Step 3: Compute the magnetic flux.
The dot product is: \[ \vec{B} \cdot \vec{A} = B A \cos \theta \] Substitute \(\theta = 90^\circ - \alpha\): \[ \cos (90^\circ - \alpha) = \sin \alpha \] However, we need to interpret the angle correctly. If the plane of the coil makes an angle \(\alpha\) with \(\vec{B}\), the angle between \(\vec{B}\) and the normal to the plane is \(\alpha\) (if \(\alpha\) is defined as the angle between the field and the plane). Thus: \[ \phi = B A \cos (90^\circ - \alpha) = B A \sin \alpha \] But in standard physics, the angle given is often between \(\vec{B}\) and the normal to the plane. Let’s correct our interpretation: if the plane makes an angle \(\alpha\) with \(\vec{B}\), the normal to the plane makes an angle \(90^\circ - \alpha\) with \(\vec{B}\), but typically, the angle \(\alpha\) in such problems is between \(\vec{B}\) and the normal. So, if \(\alpha\) is the angle between \(\vec{B}\) and the normal: \[ \phi = B A \cos \alpha \] Step 4: Match with the options.
The flux is \(BA \cos \alpha\), which matches option (B).