Question

Problem 2 The hand-turned one turn coil [N =1, (two wires Z =2N)] is placed in...

Problem 2 The hand-turned one turn coil [N =1, (two wires Z =2N)] is placed in the magnetic field of the flux density B = 0.1 T. The distance between the two conductors is 10 cm, the length of each conductor is 25 cm. If the coil is turned around the axis with the rotating speed n = 100 rev/min what is the peak voltage generated in the coil? What would be the conductors’ position in the magnetic field? Write the induced emf instantaneous value equation. ANSWER: 26.2 mV

Homework Answers

Answer #2

given

rotation speed, n = 100 rev/min

angular speed, w = 100*2*pi/60

= 10.5 rad/s

Number of turns, N = 1
Area of the loop, A = 0.1*0.25 = 0.025 m^2
B = 0.1 T

maximum induced emf in the coil, emf_max = N*A*B*w

= 1*0.025*0.1*10.5

=0.0262 V

= 26.2 mV <<<<<<<<----------------------Answer

when the loop is parallel to the magnetic field induced emf becomes maximum.

the induced emf instantaneous value equation,

emf = N*A*B*w*sin(theta)

here theta is the angle between normal to the loop and magnetic field.

answered by: anonymous
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