Question

A flux of $10^{- 3} \, Wb$ passes through a strip having an area $A=0.02 \, m^{2}$ . The plane of the strip is at an angle of $60^\circ $ to the direction of a uniform field $B$ . The value of $B$ is

• A. $0.1 \, T$
• B. $0.058 \, T$
• C. $4.0 \, \text{mT}$
• D. none of the above

Answer 1

Answer: (B)

As the angle between field and plane is $60^{o}$ , therefore angle between area vector and magnetic field is,
$\theta =30^{o}$
Magnetic flux through the strip is
$\phi=BAcos \theta $
$10^{- 3}=B\left(0.02\right)cos 30^{o}$
$B=\frac{1 0^{- 3}}{\left(0.02\right) \left(\frac{\sqrt{3}}{2}\right)}$
$B=0.058 \, T$