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




- 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.

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now