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  • PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity V as shown. The force needed to maintain constant speed of EF

PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity V as shown. The force needed to maintain constant speed of EF is.   

 

Option: 1

\frac{1}{VR}\left [ \frac{\mu _{0}IV}{2 \pi} ln \left ( \frac{b}{a} \right ) \right ]^{2}


Option: 2

\left [ \frac{\mu _{0}IV}{2\pi} ln \left ( \frac{b}{a} \right ) \right ]\frac{1}{R}


Option: 3

\left [ \frac{\mu _{0}IV}{2\pi} ln \left ( \frac{b}{a} \right ) \right ]^{2}\frac{V}{R}


Option: 4

\left [ \frac{\mu _{0}IV}{2\pi} ln \left ( \frac{b}{a} \right ) \right ]


Answers (1)

best_answer

 

Motional EMF -

\varepsilon = Blv
 

- wherein

B\rightarrow magnetic field

l\rightarrow length

u\rightarrow velocity of u perpendicular to uniform magnetic field.

 

 

Induced emf \int_{a}^{b}Bvdx = \int_{a}^{b}\frac{\mu_{0}Iv}{2 \pi }vdx

\Rightarrow Induced emf =\int_{a}^{b}Bvdx = \int_{a}^{b}\frac{\mu_{0}Iv}{2 \pi }ln\left ( \frac{b}{a} \right )

\Rightarrow power dissipated =\frac{E^{2}}{R}

Also , power = F.V \Rightarrow F = \frac{E^{2}}{VR}

\Rightarrow F =\frac{1}{VR}\left [ \frac{\mu _{0}IV}{2\pi} ln \left ( \frac{b}{a} \right ) \right ]^{2}

 

Posted by

Rakesh

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