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  • A conducting rod AC of length 4l is rotated about a point O in a uniform magnetic field B directed into the paper. AO = l and OC = 3l. Then, <img alt="" src="https://cdn.entrance360.com/m

A conducting rod AC of length 4l is rotated about a point O in a uniform magnetic field B directed into the paper. AO = l and OC = 3l. Then,

 

Option: 1

\mathrm{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{O}}=\frac{\mathrm{B} \omega \mathrm{l}^2}{2}}


Option: 2

\mathrm{\quad \mathrm{V}_{\mathrm{O}}-\mathrm{V}_{\mathrm{C}}=\frac{7}{2} \mathrm{~B} \omega l^2}


Option: 3

\mathrm{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=4 \mathrm{~B} \omega \mathrm{l}^2}


Option: 4

\mathrm{\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{O}}=\frac{9}{2} \mathrm{~B} \omega l^2}


Answers (1)

For rotating rod, induced emf, \mathrm{\mathrm{v}=\frac{1}{2} \mathrm{~B} l^2 \omega}

\mathrm{\text { For part } \mathrm{AO}, \mathrm{V}_{\mathrm{OA}}=\mathrm{V}_{\mathrm{O}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{2} \mathrm{~B} l^2 \omega}

\mathrm{\text { For part, } O C, V_{O C}=V_O-V_A=\frac{1}{2} B\left(3 l^2\right) \omega}

\mathrm{\therefore \quad V_A-V_C=4 B l^2 \omega}

Posted by

Ramraj Saini

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