A conducting rod slides down between two frictionless vertical copper tracks at

Question

A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 3.8 m/s perpendicular to a 0.55-T magnetic field. The resistance of the rod and tracks is negligible. The rod maintains electrical contact with the tracks at all times and has a length of 1.2 m. A 0.67- resistor is attached between the tops of the tracks.

(a) What is the mass of the rod?
kg

(b) Find the change in the gravitational potential energy that occurs in a time of 0.20 s.
J

(c) Find the electrical energy dissipated in the resistor in 0.20 s.
J

Additional Materials

Section 22.2

Explanation / Answer

here,

v = 3.8 m/s

B = 0.55 T

length of rod , l = 1.2 m

resistance , r = 0.67 ohm

(a)

the current in the wire , i = B*v*L /r

i = 0.55 * 3.8 * 1.2 / 0.67

i = 3.74 A

let the mass of rod be m

equating the forces

m*g = B * i * L

m * 9.8 = 0.55 * 3.74 * 1.2

m = 0.25 kg

the mass of the rod is 0.25 kg

(b)

t = 0.2 s

change in gravitational potential , PE = m * g * ( v*t)

PE = 0.25 * 9.8 * ( 3.8 * 0.2)

PE = 1.88 J

the change in gravitational potential energy is 1.88 J

(c)

electrical energy dissapated , E = i^2*r*t

E = 3.74^2 * 0.67 * 0.2

E = 1.88 J

the electrical energy dissapated is 1.88 J

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