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# A flux of $1\,mWb$ 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 $B$ . Find the value of B.A) $0.1\,T$B) $0.058\,T$C) $4.0\,mT$D) none of the above.

Last updated date: 22nd Jun 2024
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Hint:We can solve this question using the equation for calculating flux. Flux is the dot product of the magnetic field and the area. It is given as
$\phi = BA\cos \theta$
Where B is the magnetic field , A is the area and the angle $\theta$ is the angle between the area vector and the magnetic field.
By substituting the given values in this equation we can find the value of the magnetic field.

Complete step by step solution:
It is given that a flux of $1\,mWb$ passes through a strip.
Flux is a measure of the number of magnetic field lines passing through a given area. It is denoted as $\phi$
It is given by the formula
$\Rightarrow \phi = BA\cos \theta$
Where B is the magnetic field and A is the area of the cross section.
Here, $\theta$ is the angle made by the direction of the magnetic field with the area vector. Area vector is the perpendicular vector drawn at the surface of area.
Since flux is given, we can write
$\phi = 1\,mWb = 1 \times {10^{ - 3}}Wb$
Area of the strip is given as $0.02{m^2}$
$\Rightarrow A = 0.02{m^2}$
The direction of the uniform field makes an angle ${60^ \circ }$ with the strip. Thus, the angle between area vector and magnetic field will be ${90^ \circ } - {60^ \circ } = {30^ \circ }$
We need to find the value of magnetic field B .
For this let us use the equation for calculating flux.
$\phi = BA\cos \theta$
On substituting the given values we get
$1 \times {10^{ - 3}} = B \times 0.02 \times \cos \,{30^0}$
$\Rightarrow B = \dfrac{{{{10}^{ - 3}}}}{{2 \times {{10}^{ - 2}} \times \dfrac{{\sqrt 3 }}{2}}}$
$\therefore B = 0.058\,T$.
This is the value of the magnetic field.

So, the correct answer is option B.

Note: Remember that the angle $\theta$ in the equation for calculating flux is the angle between area vector and the magnetic field. An area vector is a normal vector drawn at the surface of a plane. The given angle ${60^ \circ }$ is the angle made by the plane with the field. So subtract this angle from ${90^ \circ }$ to get the angle made by the area vector with the magnetic field.