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).The magnetic field is directed along the positive $X$-axis. The top surface is inclined at an angle of $30^{\circ}$ with the horizontal. The normal to this surface makes an angle of $	heta = 90^{\circ} - 30^{\circ} = 60^{\circ}$ with the $X$-axis.The dimensions of the top surface are $10 \, cm 	imes 10 \, cm = 0.1 \, m 	imes 0.1 \, m = 0.01 \, m^2$.Substituting the values:$\phi = (0.2 \, T) 	imes (0.01 \, m^2) 	imes \cos(60^{\circ})$$\phi = 0.002 	imes 0.5 = 0.001 \, Wb$Since $1 \, Wb = 1000 \, m-Wb$,we have:$\phi = 0.001 	imes 1000 = 1 \, m-Wb$.

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

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