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(B) The magnetic flux $\Phi$ is given by the formula $\Phi = B A \cos(	heta)$,where $	heta$ is the angle between the magnetic field vector $\vec{B}$

Vedclass · Accepted Answer

$0.058 \ T$

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(B) The magnetic flux $\Phi$ is given by the formula $\Phi = B A \cos(	heta)$,where $	heta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$.Given that the plane of the strip makes an angle of $60^{\circ}$ with the magnetic field,the angle $	heta$ between the area vector (which is perpendicular to the plane) and the magnetic field is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.Given values: $\Phi = 1 \ mWb = 10^{-3} \ Wb$,$A = 0.02 \ m^2$,and $	heta = 30^{\circ}$.Substituting these values into the formula: $10^{-3} = B 	imes 0.02 	imes \cos(30^{\circ})$.Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866$,we have $10^{-3} = B 	imes 0.02 	imes 0.866$.$B = \frac{10^{-3}}{0.02 	imes 0.866} = \frac{0.001}{0.01732} \approx 0.0577 \ T$.Rounding this value gives $B \approx 0.058 \ T$.

$A$ 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 $B$. The value of $B$ is

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