Question

What is the magnitude of the force on a current carrying conductor of length L in a perpendicular magnetic field B?

(a) F=\(\frac {I}{lB}\)

(b) F=IlB^2

(c) F=I^2  lB

(d) F=IlB

The question was posed to me in homework.

My doubt is from Energy Consideration: A Quantitative Study topic in section Electromagnetic Induction of Physics – Class 12

Answer 1

Right choice is (d) F=IlB

The best explanation: The current passing through conductor in a perpendicular magnetic field is given as:

I=\(\frac {\varepsilon}{r}\)

I=\(\frac {Blv}{r}\)

Since it is a perpendicular magnetic field, the force ➔ I(l×B) is directed outwards in the direction opposite to the velocity of the rod. The magnitude of the force is given as:

F=IlB

We can also write it as ➔ F=IlB=\(\frac {B^2 l^2 v}{r}\)