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A uniform horizontal magnetic field B exists in the region A B C D. A rectangular loop of mass m and horizontal side l and resistance R is placed in the magnetic field as shown in figure. With what velocity should it be pushed down so that it continues to falls without acceleration?
                       

Option: 1

\frac{\mathrm{mgR}}{\mathrm{B}^2 1^2}


Option: 2

\frac{\mathrm{B}^2 \mathrm{l}^2}{\mathrm{mgR}}


Option: 3

\frac{\mathrm{mg}}{\mathrm{BIR}}


Option: 4

\frac{\mathrm{RB} 1}{\mathrm{mg}}


Answers (1)

best_answer

According to Lenz’s law, the magnetic force is given by
  \mathrm{F}=\frac{\mathrm{B}^2 \mathrm{l}^2 \mathrm{~V}}{\mathrm{R}}
According to given problem F should be equal to m g.
 \text { Hence } \mathrm{mg}=\frac{\mathrm{B}^2 \mathrm{l}^2 \mathrm{v}}{\mathrm{R}} \text { or } \mathrm{v}=\frac{\mathrm{mgR}}{\mathrm{B}^2 \mathrm{l}^2}

Posted by

Ritika Harsh

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