A wire cd of length l , mass m, is sliding without friction on conducting rails ax and by as shown in figure. The vertical rails are connected to one another via an external resistance R. The entire circuit is placed in a region of space having a uniform magnetic field B. The field is ⊥ to the plane of circuit & directed outwards. The steady speed of rod cd is

  • Option 1)

    \frac{mg R}{Bl}

  • Option 2)

    mg \frac{R}{B^{2}l^{2}}

  • Option 3)

    mg \frac{R}{Bl^{2}}

  • Option 4)

    mg \frac{R}{B^{2}l}

 

Answers (1)

 

Magnetic force -

F=Ilb=B\left ( \frac{Blv}{R} \right )l

F=\frac{B^{2}vl^{2}}{R}

 

 

- wherein

B\rightarrow Magnetic field

R\rightarrow Resistance

V\rightarrow Velocity

 

 In steady state F_{net}=0

\Rightarrow magnetic force F_{mag} = mg

\Rightarrow IlB=mg

\because I=\frac{\varepsilon }{R} = \frac{Blv}{R}\Rightarrow \frac{B^{2}l^{2}}{R} v = mg

\Rightarrow v \frac{mgR}{B^{2}l_{2}}


Option 1)

\frac{mg R}{Bl}

This solution is incorrect.

Option 2)

mg \frac{R}{B^{2}l^{2}}

This solution is correct.

Option 3)

mg \frac{R}{Bl^{2}}

This solution is incorrect.

Option 4)

mg \frac{R}{B^{2}l}

This solution is incorrect.

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