Question

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

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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