Permanent magnetismMedium
Question
A magnet 10 cm long and having a pole strength 2 amp m is deflected through 30o from the magnetic meridian. The horizontal component of earth′s induction is 0.32 × 10-4 tesla then the value of deflecting couple is :
Options
A.32 × 10-7 Nm
B.16 × 10-7 Nm
C.64 × 10-7 Nm
D.48 × 10-7 Nm
Solution
M = 2 × 
= 0.2
Value of restoring couple = MH sin φ
= 0.2 × 0.32 × 10-4 sin 30o
= 0.2 × 0.32 × 10-4 ×
= 32 × 10-7 Nm
= 0.2Value of restoring couple = MH sin φ
= 0.2 × 0.32 × 10-4 sin 30o
= 0.2 × 0.32 × 10-4 ×
= 32 × 10-7 NmCreate a free account to view solution
View Solution FreeMore Permanent magnetism Questions
Magnetic field lines produced by a bar magnet, cuts...The magnetic moment of a straight wire of length L is M. If the wire is bent in form of a semicirle then its magnetic mo...A short bar magnet of magnetic moment 0.4 JT-1 is placed in a formula in a uniform magnetic field of 0.16 T. The magnet ...At any place horizontal and vertical component of earth magnetic field are equal then angle of dip at that place :-...A vibration magnetometer consists of two identcial bar magnets placed one over the other such that they are perpendicula...