Question

A conductor of length 50 cm carrying of 5 a is placed perpendicular to a magnetic field of induction 2 x 10-3 t. Find the force on the conductor.

Answer 1

GOOD MORNING

F=BILsin(theta), Let it be Equation 1

Where, B=Field of Induction=2* 10^-3

              I=Current=5 A ; L= Length=50cm ; sin(theta) =90 (as it is held perpendicular)

Applying all the values, in Equation 1  

F=BIL sin90

F=500 *10^-3

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