A closely wound solenoid of 100 turns and area of cross section 5x10^-6 m^2 carries a current of 5A. What is the magnetic field associated with it?
Answers
Answer:
A closely wound solenoid of 800 turns and area of cross-section 2.5 x 10–4 m2 carries a current of 3.0 A. ... Magnetic field is developed along the axis of the solenoid. Therefore, solenoid acts like a bar magnet. The direction is determined by the sense of flow of the current.
Answer:
a)Given,
Number of turns of coil, N=2000
Area of cross-section of solenoid, A=1.6×10
−4 m 2
Current passing through the coil,I=4A
Magnetic moment, M=NIA
=2000×4×1.6×10
−4 =1.28Am
2
The direction of
M
is along the axis of the solenoid in the direction in the related to the sense of current via the right-handed screw rule.
b) Uniform magnetic field applies, B=7.5×10
−2 T
The angle between the axis of the solenoid and magnetic field θ=30
0
Since the magnetic field is uniform on the solenoid, the force acting on the solenoid is 0.
Torque is given by,
τ=MBsinθ=1.28×7.5×10 −2
sin30 0
=1.28×7.5×10 −2 × 21 J=0.048J
The direction of the torque is such that the solenoid tends to align the axis of the solenoid (magnetic moment vector) along the direction of the magnetic field
B