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  • A conducting rod of length of l =  slides at constant velocity 'v' on two parallel conducting rails, placed in a uniform and constant magnetic field B perpendicular to the plane of the

A conducting rod of length of l =  slides at constant velocity 'v' on two parallel conducting rails, placed in a uniform and constant magnetic field B perpendicular to the plane of the rails as shown in the figure. A resistance R is connected between the two ends of the rail. Find the electric power dissipated in the resistor.

 

 

Option: 1

P=\frac{Bl^{2} v^{2}}{R}


Option: 2

P=\frac{B^{2} l^{2} v^{2}}{R}


Option: 3

P=\frac{B^{2} l^{2} v}{R}


Option: 4

P=\frac{B^{2} l v^{2}}{R}


Answers (1)

best_answer

 

 

\begin{array}{l}{\text { induced emf, }|\varepsilon|=\left|-\frac{d \phi}{d t}\right|=B v l} \\ \\ {\text { induced current, } i=\frac{\varepsilon}{R}=\frac{B v l}{R}} \\ \\ So, \\ {\qquad F =B\left(\frac{B v l}{R}\right) l} \\\\ {F =\left(\frac{B^{2} v l^{2}}{R}\right)}\end{array}

\begin{array}{l}{\text { Power dissipitated } P=F \cdot v}\\ \\ {P=\left(\frac{B^{2} v l^{2}}{R}\right) \cdot v} \\ \\ {P=\frac{B^{2} v^{2} l^{2}}{R}}\end{array}

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sudhir.kumar

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