A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwis

Question

A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the ground, and the Earth's magnetic field points due north, has a magnitude of 5.8 times 10^-5 T, and makes a downward angle of 25 degree with the vertical, what is the torque on the loop? A) 1.7 times 10^-2 N middot m west B) 3.6 times 10^-2 N middot m west C) 1.7 times 10^2 N middot m east D) 3.6 times 10^-2 N middot m east E) 3.6 times 10^-2 N middot m south The diagrams show five possible orientations of a magnetic dipole mu vector in a uniform magnetic field B vector. For which of these is the potential energy the greatest? A) I B) II C) III D) IV E) V

Explanation / Answer

9)

Radius of the coil , R= 12 cm = 0.12 m
Current flowing in the coil, I = 15 A
Number of turns on the coil, n = 1000

Angle made by the plane of the coil with magnetic field, = 25°
Strength of magnetic field, B = 5.8 x 10-5 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
= n BIA sin
Where,
A = Area of the circular coil = R2 = 3.14 x 0.12 x 0.12 = 0.045 m2
= 1000 × 5.8 x 10-5 × 15 x 0.045 × 0.42
= 1.7 x 10-2 N m
Hence, the magnitude of the torque experienced by the coil is 1.7 x 10-2 N m

Option is C

Since current is carried counter clockwise

10) Potential Energy is given by U = -pE cos

Since cos is maximum when mu and E are in the same direction i.e., = 0 but in the negative sense

Hence Option E is the right answer.

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