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  • Confused! kindly explain, A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of θ with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic f

A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of θ with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic field B normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is :

  • Option 1)

    \frac{mg R tan\Theta }{B^{2}I^{2}}

  • Option 2)

    \frac{mg R cot\Theta }{B^{2}I^{2}}

  • Option 3)

    \frac{mg R sin\Theta }{B^{2}I^{2}}

  • Option 4)

    \frac{mg R cos\Theta }{B^{2}I^{2}}

 

Answers (2)

best_answer

As we learnt @3193

At termnal Velocity magnetic force =mgsin\theta

EMF=vlB

I=\frac{vlB}{R}

\therefore magnetic force =IlB=\frac{vl^{2}B^{2}}{R}

\frac{vl^{2}B^{2}}{R}=mgsin\theta

\therefore v= \frac{mgRsin\theta}{l^{2}B^{2}}


Option 1)

\frac{mg R tan\Theta }{B^{2}I^{2}}

Option 2)

\frac{mg R cot\Theta }{B^{2}I^{2}}

Option 3)

\frac{mg R sin\Theta }{B^{2}I^{2}}

Option 4)

\frac{mg R cos\Theta }{B^{2}I^{2}}

Posted by

Avinash

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