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Force on a current carrying wire placed in a magnetic field

Question

A rectangular loop with a sliding connector of length l=1.0 m is situated in a uniform magnetic field B = 2 T perpendicular to the plane of loop. Resistance of connector is r = 2 Ω. Two resistances of 6 Ω and 3 Ω are connected as shown in figure. The external force required to keep the connector moving with a constant velocity v = 2 m/s is
 

Moderate
Solution

Motional emf  

e = Bvl \Rightarrow e = 2 \times 2 \times 1 = 4\;V


This acts as a cell of emf  E = 4 V and internal resistance  r = 2Ω .
This simple circuit can be drawn as follows


Current through the connector  

i = \frac{4}{{2 + 2}} = 1\;A


 

\therefore

 magnetic force on connector 

{F_m} = Bil = 2 \times 1 \times 1 = 2\;N


            (Towards left)

Hence, the external force should be 2 N towards right.



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Two infinitely current carrying insulated wires AB and CD are carrying same current  I. The wires cross each other at right angles as shown in figure. Three identical conductors P1Q1 , P2Q2 and P3Q3, carrying same current are placed a shown in figure and the forces acting on them are F1 , F2 and F3 respectively. Then select the correct option.


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