Question

A circular disc of radius 20 cm is placed in uniform magnetic field of induction \(\frac{7}{22} \text{Wb/m}^2\) in such a way that its axis makes an angle of \(60^{\circ}\) with \(\overset{⃗}{B}\). The magnetic flux linked with the disc is \((cos60^{\circ} = 0.5)\)

• A. \(0.01\) Wb
• B. \(0.02\) Wb
• C. \(0.06\) Wb
• D. \(0.08\) Wb

Hint:

Flux is B A cos(theta) with theta measured from the normal (axis).

Answer 1

The Correct Option is B

Step 1: Approach
Work in steps with simple numbers.

Step 2: Steps
$B\times\pi r^2=\dfrac7{22}\times\dfrac{22}{7}\times0.04=0.04$ Wb for a disc facing the field. Tilting the axis by $60^\circ$ cuts this to half, since $\cos60^\circ=0.5$.

Step 3: Answer
$0.02$ Wb, option (B).

Final Answer:
The flux is B times area times cos 60, which is 0.04 times 0.5 = 0.02 Wb, option (B). \[ \boxed{0.02\ \text{Wb}} \]