A horizontal rod of mass 10 gm and length 10 cm is placed on a smooth plane inclined at an angle of 60 degree with the horizontal, with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field of induction B is applied vertically downwards. If the current through the rod is 1.73 ampere, then the value of B for which the rod remains stationary on the inclined plane is

  • Option 1)

    1.73 Tesla

  • Option 2)

    1 / 1.73Tesla

  • Option 3)

    1 Tesla


  • Option 4)

    None of the above


Answers (1)

As we learnt in

Total magnetic force -

\underset{F}{\rightarrow}=i ( \underset{L}{\rightarrow}\times \underset{B}{\rightarrow})

- wherein




For equilibrium

mg sin \Theta = i l B cos \Theta

 \Rightarrow 10^{-2} \times 10 \times\frac{\sqrt{3}}{2}=1.73 \times 0.1 \times B \times\frac{1}{2}

\Rightarrow    B = 1 T

Option 1)

1.73 Tesla


Option 2)

1 / 1.73Tesla


Option 3)

1 Tesla



Option 4)

None of the above


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