Find the magnetic flux density in Wb/m^2,if a | vedclass

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

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(B) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos 	heta$,where $	heta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is normal to the surface).From the figure,the angle between the surface and the vertical is $60^\circ$. The area vector $\vec{A}$ is perpendicular to the surface,so the angle between the horizontal magnetic field $\vec{B}$ and the area vector $\vec{A}$ is $	heta = 60^\circ$.Given: $\phi = 2 	imes 10^{-3} \ Wb$,$A = 10 \ cm^2 = 10 	imes 10^{-4} \ m^2 = 10^{-3} \ m^2$,and $	heta = 60^\circ$.Using $\phi = BA \cos 	heta$,we have $B = \frac{\phi}{A \cos 	heta}$.Substituting the values: $B = \frac{2 	imes 10^{-3}}{(10^{-3}) \cos 60^\circ} = \frac{2 	imes 10^{-3}}{(10^{-3}) 	imes (1/2)} = \frac{2}{1/2} = 4 \ Wb/m^2$.

Find the magnetic flux density in $Wb/m^2$,if a magnetic flux of $2 \times 10^{-3} \ Wb$ passes through a slant area $A = 10 \ cm^2$ as shown in the figure.

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