A circular disc of radius 0.2 ~m is placed in a uniform magnetic field of induction (1)/(π) ((W b)/(m^2)) in such a way that its axis makes an angle of 60^ with B . The magnetic flux linked with the disc is

Correct answer: A. 0.02 ~Wb The magnetic flux φ passing surface of area A placed in a u field B is given by φ=B A θ where θ is the angle between aligned & Here, θ=60^ , B=(1)/(π) Wb / m^2, A=π(0.2)^2 & Therefore, φ=(1)/(π) × π(0.2)^2 × 60^ & =(0.2)^2 × (1)/(2) & =0.02 Wb & aligned

NEET 2008

Exam: NEET
Year: 2008
Subject: Physics
Chapter: Electromagnetic Induction

NEET 2008 · Physics · Electromagnetic Induction

A circular disc of radius 0.2 ~m is placed in a uniform magnetic...

A circular disc of radius

0.2 \mathrm{~m}

is placed in a uniform magnetic field of induction

\frac{1}{\pi}

\left(\frac{W b}{m^2}\right)

in such a way that its axis makes an angle of

60^{\circ} with

\overrightarrow{\mathbf{B}}

. The magnetic flux linked with the disc is

Options

A $0.02 \mathrm{~Wb}$
B $0.06 \mathrm{~Wb}$
C $0.08 \mathrm{~Wb}$
D $0.01 \mathrm{~Wb}$

Correct answer

A. $0.02 \mathrm{~Wb}$

Step-by-step solution

The magnetic flux

\phi

passing surface of area A

placed in a u field

is given by

\phi=B A \cos \theta

where

\theta is the

angle between

\begin{aligned} & \text{Here, } \theta=60^{\circ}, B=\frac{1}{\pi} \mathrm{Wb} / \mathrm{m}^2, A=\pi(0.2)^2 \\ & \text{Therefore, } \phi=\frac{1}{\pi} \times \pi(0.2)^2 \times \cos 60^{\circ} \\ & =(0.2)^2 \times \frac{1}{2} \\ & =0.02 \mathrm{ Wb} \\ & \end{aligned}

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