Physics-
General
Easy

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 equals 0.15 t e s l a. If the resistance is 3 capital omega, force needed to move the rod as indicated with a constant speed of 2 m divided by s e c is

  1. 3.75 cross times 1 0 to the power of negative 3 end exponentN    
  2. 3.75 cross times 1 0 to the power of negative 2 end exponent N    
  3. 3.75 cross times 1 0 to the power of 2 end exponent N    
  4. 3.75 cross times 1 0 to the power of negative 4 end exponent N    

The correct answer is: 3.75 cross times 1 0 to the power of negative 3 end exponentN


    Induced current in the circuit i equals fraction numerator B v l over denominator R end fraction
    Magnetic force acting on the wire F subscript m end subscript equals B i l equals B open parentheses fraction numerator B v l over denominator R end fraction close parentheses   l
    rightwards double arrow F subscript m end subscript equals fraction numerator B to the power of 2 end exponent v l to the power of 2 end exponent over denominator R end fraction External force needed to move the rod with constant velocity
    left parenthesis F subscript m end subscript right parenthesis equals fraction numerator B to the power of 2 end exponent v l to the power of 2 end exponent over denominator R end fraction equals fraction numerator left parenthesis 0.15 right parenthesis to the power of 2 end exponent cross times left parenthesis 2 right parenthesis cross times left parenthesis 0.5 right parenthesis to the power of 2 end exponent over denominator 3 end fraction equals 3.75 cross times 1 0 to the power of negative 3 end exponent N

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