Magnetic field due to currentHard
Question
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60. The torque needed to maintain the needle in this position will be
Options
A.
B.2√3 J
C.3 J
D.√3 J
Solution
Work done Uf - Ui = - MBcos60o - (-MBcos0o)
= - MB
MB = 2√3
= MB sin 60o = 
= - MB
MB = 2√3 = MB sin 60o = Create a free account to view solution
View Solution FreeMore Magnetic field due to current Questions
A particle of specified charge (q/m) is projected from the origin of coordinates with initial velocity [uî + vĵ...The field produced by a moving charged particle is :-...The unit of magnetic moment will be :-...A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved, by rotating i...A square coil of side a carries a current I. The magnetic field at the center of the coil is...