Question

MCQ (100)
Q43. A current carrying conductor placed perpendicular
to the direction of magnetic field experiences a
force of magnitude equal to :
(A) BIL
(B) 0
(C) 1
(D) 2.​

Answer 1

(Option1)

The force of a current carrying conductor is given as:

F = BIL×sin∅

where,

B is magnetic flux density

I is current in conductor

L is length of the conductor

∅ is the angle between the magnetic field and current

Since ∅ is given as 90°,

F = BIL×sin90

(sin 90 = 1)

F = BIL

Hence, the force of magnitude between a current carrying conductor and magnetic field with 90° angle is equal to BIL.