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Force on a current carrying wire placed in a magnetic field

Question

As shown in the figure a metal rod makes contact and complete the circuit. The circuit is perpendicular to the magnetic field with B = 0.15\,tesla.  If the resistance is 3Ω , force needed to move the rod as indicated with a constant speed of 2m/sec  is         
   

Moderate
Solution

Induced current in the circuit  i = \frac{{Bvl}}{R}
Magnetic force acting on the wire  {F_m} = Bil = B\left( {\frac{{Bvl}}{R}} \right)\;l
\Rightarrow {F_m} = \frac{{{B^2}v{l^2}}}{R}  External force needed to move the rod with constant velocity       
({F_m}) = \frac{{{B^2}v{l^2}}}{R} = \frac{{{{(0.15)}^2} \times (2) \times {{(0.5)}^2}}}{3} = 3.75 \times {10^{ - 3}}N

 



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