Magnetic field due to currentHard
Question
A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through 60o. The torque needed to maintain the needle in this position will be -
Options
A.√3 W
B.W
C.(√3 / 2) W
D.2W
Solution
W = MB(1-cos 60o) = 
τ = MB sin 60o = √3
= √3 W
τ = MB sin 60o = √3
Create a free account to view solution
View Solution FreeMore Magnetic field due to current Questions
A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields and with a velocity...A metal wire of mass ′m′ slides without friction on two rails spaced at a distance ′d′ apart. Th...A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinderis placed coaxi...A disc of radius r and positive charge q is rotating with angular speed ω in a uniform magnetic field B about a fix...Radius of current carrying coil is ′R′. Then ratio of magnetic fields at the centre of the coil to the axial...