A current carrying rod of length 2 m is kept at an angle of 45 degrees with the magnetic field 0.5 T and carrying a current of 1.4 A. What will be the force experienced by the charge carriers?
Answer & explanation
Correct answer: option 1
F = IlB Sin θ
I = 1.4 A ; l = 2 m ; B = 0.5 T ; θ = 45
F = 1.4x2x0.5x$\frac{1}{\sqrt 2} = 1N $