Get Answers to all your Questions

header-bg qa

A circular loop of radius 1m is kept in a magnetic field of strength 2 T directed perpendicular to the plane of loop. Resistance of the loop wire is 2/π Ω/m. A conductor of length 2 m is sliding with a speed 1 m/s as shown in the figure. Find the instantaneous force acting on the rod [Assume that the rod has negligible resistance]

  • Option 1)

    8 N

  • Option 2)

    16 N

  • Option 3)

    32 N

  • Option 4)

    64 N


Answers (1)


Induced Current -

I= \frac{\varepsilon }{R}= \frac{Blv}{R}



 Given figure can be redrawn as:

 Electromagnetic Induction

R=\frac{2}{\pi }\times \pi r=\frac{2}{\pi }\times \pi \times 1 =2 ohm

2 ohm\: and\: 2ohm\: are\:in\: parallel

R_eq= 1ohm

= Blv- B.(2r)v

= 2T\times 2\times1\times1=4V

I=\frac{\varepsilon }{R_{eq}}=\frac{4}{1} =4A

\therefore Force on wire = I. l.B

I.(2r)B)=4\times 2\times 1\times 2 =16N

Option 1)

8 N


Option 2)

16 N


Option 3)

32 N


Option 4)

64 N


Posted by


View full answer