A metal rod with a mass of 0.30 kg rolls down an inclined plane (inclined at 10

Question

A metal rod with a mass of 0.30 kg rolls down an inclined plane (inclined at 10 degrees above the horizontal). The rod makes contact with two metal rails that are connected to a R = .15 ohm resistor making a closed circuit. The rails are separated from one another by d = 0.12 m. The entire circuit is in a uniform magnetic field of B = 2.5 T that is directed vertically. What is the power dissipated in the resistor when the rod readies terminal velocity? (Neglect sliding and rolling friction of the rod on the rails.)

Explanation / Answer

Given that

mass m=.30 kg

resistance R=0.15 ohm

distance l=0.12 m

magnetic field B=2.5 T

now we find thge current

Bil=mgsin(theta)

2.5*i*0.12=0.3*9.8*sin(10)

current i=0.51/0.3=1.7 A

now we find the power dispated

power dispated P=1.7^2*0.15=0.434 W

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