At one instant a proton (charge = 16 x 10^10 C) is moving in the x direction wit

Question

At one instant a proton (charge = 16 x 10^10 C) is moving in the x direction with a velocity of v = 3 x 10^5 m/s. A magnetic field of 0.6 T is applied in they direction to this moving charge. Calculate the direction and magnitude of the magnetic force on the proton. Calculate the direction and magnitude of the magnetic force on the electron under the same situation as in part. An ac generator producing 10 V (rms) at 45 Hertz frequency is connected series with a 30 - ohms resistor, a 200 - mH inductor, and a 100 - microF capaotor. Calculate The impedance in the circuit The rms current The power factor Page 3 of 7

Explanation / Answer

6) given data

q = 1.6*10^-19 C

v = 3*10^5 m/s

B = 0.6 T

a) we know, F = q*v*B*sin(theta)

= 1.6*10^-19*3*10^5*0.6*sin(90)

= 2.88*10^-15 N

direction : towards +z axis

b) magnitude will be same.

F = 2.88*10^-15 N

Direction : towards -z axis

7)

given

R = 30 ohms

C = 100 micro F

L = 200 mH

f = 45 hz

Vrms = 10 v

a)
capacitive reactance, XC = 1/(2*pi*f*C) = 1/(2*pi*45*100*10^-6) = 35.37 ohms

Inductive reactance, XL = 2*pi*f*L = 2*pi*45*200*10^-3 = 56.55 ohms

impedance, z = sqrt(R^2 + (XL - XC)^2)

= sqrt(30^2 + (56.55 - 35.37)^2)

= 36.7 ohms

b) Irms = Vrms/z

= 10/36.7

= 0.272 A

c) power factor, cos(phi) = R/z

= 30/36.7

= 0.817

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