A semicircle conducting ring of radius R is p | vedclass

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

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All of the above

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(D) The magnetic flux $\phi$ through a surface is given by $\phi = \vec{B} \cdot \vec{A} = BA \cos 	heta$,where $	heta$ is the angle between the magnetic field $\vec{B}$ and the area vector $\vec{A}$.In this case,the ring lies in the $xy$ plane,so its area vector $\vec{A}$ points along the $z$-axis. The magnetic field $\vec{B}$ is along the $x$-axis.Since $\vec{B}$ is along the $x$-axis and $\vec{A}$ is along the $z$-axis,the angle $	heta = 90^\circ$. Therefore,the magnetic flux $\phi = BA \cos(90^\circ) = 0$.According to Faraday's law,the induced $emf$ is $E = -\frac{d\phi}{dt}$.If the ring moves along the $x$,$y$,or $z$ axis,the magnetic field remains uniform and the orientation of the ring relative to the field does not change. Thus,the flux remains zero at all times,and $\frac{d\phi}{dt} = 0$.Consequently,no $emf$ is induced in the ring regardless of whether it moves along the $x$,$y$,or $z$ axis. Therefore,all the given options are correct.

$A$ semicircle conducting ring of radius $R$ is placed in the $xy$ plane,as shown in the figure. $A$ uniform magnetic field is set up along the $x$-axis. No $emf$ will be induced in the ring if:

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