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: Understanding the Concept
Magnetic flux is \(\Phi=BA\cos\theta\), where \(\theta\) is the angle between \(\vec B\) and the normal to the surface. The axis of the disc is its normal.

Step 2: Area
\[ A=\pi r^2=\frac{22}{7}\times(0.2)^2=\frac{22}{7}\times0.04 \]

Step 3: Flux
\[ \Phi=\frac7{22}\times\frac{22}{7}\times0.04\times\cos60^\circ=0.04\times0.5=0.02\ \text{Wb} \]

Step 4: Check the options
The value 0.04 Wb would result if the angle were zero. The value 0.01 Wb would come from also multiplying by 0.5 twice. Our result is 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). \[ \boxed{0.02\ \text{Wb}} \]