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Q. A flux of $1 \, mWb$ passes through a strip having an area $A=\text{0.02} \, m^{2}$ . 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

NTA AbhyasNTA Abhyas 2020

Solution:

We know that, $\phi=BAcos \theta $
Angle between $\overset{ \rightarrow }{A}$ and $\overset{ \rightarrow }{B}$ is $30^\circ $ .
$\phi=1\times 10^{- 3} \, Wb$
$A=0.02m^{2}$
$\Rightarrow 10^{- 3}=B\times 2\times 10^{- 2}\times cos 30^{o}$
$\Rightarrow B=0.058 \, T$