PART II Group electric resistance of the rails can be neglected as well as the f

Question

PART II Group electric resistance of the rails can be neglected as well as the friction. The Circuit is in a external 0.263 m, R-255 , and B = 0.890 T. a) Is clockwise or counterclockwise the direction of the induced current? b) Find the e) Determine the Force which is being applied to the right to maintain the rod moving with constant PI. The conducting bar makes contact with metal circuit which composed of rails and a resistor. The uniform ma magnetic field and the bar is moving to the right with constant speed v. The bar has length bar when it is moving toward the right with a speed 7.50 m/s speed. P2. A series ac circuit contains a 250 Resistor, a 10.0 mH Inductor, a 3.50 F Capacitor, and an AC power source of voltage amplitude 35.0 V operating at an angular frequency of 260 rad/s. Calculate, a) The Impedance of the circuit. b) The total current amplitude. b) The Power factor Group 2 P3. A sinusoidal electromagnetic wave having a Magnetic field of amplitude 2.30#T and a wavelength of 532 nm is traveling in the +x direction through a medium with diffraction index n = 1.6. Assume that the Absolute permeability of the medium is -20 H m (remember that H(Henry)-T m2/A) a) Calculate the velocity of the wave. b) Calculate Intensity of the wave. P4. An electromagnetic wave with frequency 55.0Hz travels in an insulating magnetic material that has dielectric constant 3.31 and relative permeability 6.23 at this frequency. The Electric field has amplitude 7.20*mN/C. a) Calculate the wavelength of the wave c) Calculate the amplitude of the Magnetic field.

Explanation / Answer

PART II :

P1)

The answer to this part will depend on the direction of applied magnetic field B = 0.89 T

If the magnetic field is into the plane of the loop, the induced current will be counterclockwise and if the magnetic field is out of the loop, the current will be clockwise <---- Applying Lenz's Law

P2)

induced emf = B*L*v = 1.76 V <-----------answer

c)

So, induced current, I = BLv/R

= 0.89*0.263*7.5/25.5

= 0.069 A

Force = B*I*L = 0.89*0.069*0.263

= 0.016 N

P2)

a)

Impedance , Z = sqrt(R^2 + (W*L - 1/(W*C))^2)

= sqrt(250^2 + (260*0.01 - 1/(260*3.5*10^-6))^2)

= 1124.4 ohm

b)

Current = 35/1124.4

= 0.031 A

c)

Power factor = R/Z = 250/1124.4

= 0.222

P3)

a)

velocity of wave = c/n

= 3*10^8/1.6

= 1.875*10^8 m/s

b)

Intensity of wave = B/u = 2.3*10^-6/20

= 1.15*10^-7

P4)

a)

Wavelength = frequency/v

= 55/(3*10^8/sqrt(3.31*6.23))

= 8.33*10^-7 m

b)

Magnetic field amplitude = E/v

= 7.2*10^-3/(3*10^8/sqrt(3.31*6.23))

= 1.09*10^-10 T

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