Consider the following figure. A uniform magn | vedclass

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

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(C) The magnetic flux $\phi$ is given by $\phi = \vec{B} \cdot \vec{A} = B A \cos 	heta$,where $	heta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (normal to the surface).Given: Magnetic field $B = 0.2 \, T$ along the positive $x-$axis.The top surface is a rectangle with dimensions $10 \, cm 	imes 10 \, cm = 0.1 \, m 	imes 0.1 \, m = 0.01 \, m^2$.The normal to the top surface makes an angle of $60^{\circ}$ with the vertical,which means it makes an angle of $90^{\circ} - 60^{\circ} = 30^{\circ}$ with the $x-$axis (or based on the geometry,the angle $	heta$ between the normal and the $x-$axis is $60^{\circ}$ as given in the diagram).Using $	heta = 60^{\circ}$:$\phi = B A \cos 60^{\circ} = 0.2 	imes 0.01 	imes \cos 60^{\circ} = 0.2 	imes 0.01 	imes 0.5 = 0.001 \, Wb$.Converting to $m-Wb$: $0.001 \, Wb = 1 \, m-Wb$.

Consider the following figure. $A$ uniform magnetic field of $0.2 \, T$ is directed along the positive $x-$axis. What is the magnetic flux through the top surface of the figure in $m-Wb$?

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