Q114 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
A circular disc of radius 20 cm is placed in uniform magnetic field of induction $\dfrac{7}{22}\ \text{Wb/m}^2$ in such a way that its axis makes an angle of $60^\circ$ with $\vec{B}$. The magnetic flux linked with the disc is $(\cos 60^\circ = 0.5)$
- A$0.01$ Wb
- Correct answerB$0.02$ Wb
- C$0.06$ Wb
- D$0.08$ Wb
Solution
Radius of the disc, $r = 20 \text{ cm} = 0.2 \text{ m}$
Area of the disc, $A = \pi r^2 = \dfrac{22}{7} \times (0.2)^2 = \dfrac{22}{7} \times 0.04 \text{ m}^2$
Magnetic field, $B = \dfrac{7}{22} \text{ Wb/m}^2$
Angle between the axis of the disc (area vector) and the magnetic field, $\theta = 60^\circ$
Magnetic flux linked with the disc is given by $\Phi = B A \cos \theta$
Substituting the values:
$\Phi = \dfrac{7}{22} \times \left( \dfrac{22}{7} \times 0.04 \right) \times \cos 60^\circ$
$\Phi = 0.04 \times 0.5 = 0.02 \text{ Wb}$
Answer: $0.02$ Wb
Area of the disc, $A = \pi r^2 = \dfrac{22}{7} \times (0.2)^2 = \dfrac{22}{7} \times 0.04 \text{ m}^2$
Magnetic field, $B = \dfrac{7}{22} \text{ Wb/m}^2$
Angle between the axis of the disc (area vector) and the magnetic field, $\theta = 60^\circ$
Magnetic flux linked with the disc is given by $\Phi = B A \cos \theta$
Substituting the values:
$\Phi = \dfrac{7}{22} \times \left( \dfrac{22}{7} \times 0.04 \right) \times \cos 60^\circ$
$\Phi = 0.04 \times 0.5 = 0.02 \text{ Wb}$
Answer: $0.02$ Wb
Hence, the correct answer is option: 'B'.