Question

My Notes Ask Your Teacher 3. -13 points SerPSE9 31.P.029. A conducting rod of length moves on two horizontal, frictionless rails, as in the figure. A constant force of 1.00 N moves the bar at 1.00 m/s through a magnetic field B that is directed into the monitor (a) What is the current through the 7.00 resistor R? (b) What is the rate at which energy is delivered to the resistor? (c) What is the mechanical power delivered by the force Fop?

Answer 1

a.)

Since velocity of rod is constant.

So, magnetic force on rod is given by,

Fb = B*i*l = Constant force on rod

also voltage across rod will be

E = B*v*l

from ohm's law,

E = i*R

So, Fb = i^2*R/v

i = sqrt(Fb*v/R)

given values,

Fb = 1.00 N

v = 1.00 m/sec.

R = 7.00 ohm

So, i = sqrt(1*1/7)

i = 0.378 Amp

b.)

Rate of energy = Power = i^2*R

P = (0.378^2)*7.00

P = 1.00 W

c.)

Mechanical power is given by,

P = F*v

from given values,

P = 1.00*1.00

P = 1.00 W

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