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 4.3 m/s perpendicular to a 0.56-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 1.5 m. A 0.82-? resistor is attached between the tops of the tracks.

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


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

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

Can you explain please?

A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.3 m/s perpendicular to a 0.56-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 1.5 m. A 0.82-? resistor is attached between the tops of the tracks. (a) What is the mass of the rod in kg? (b) Find the change in the gravitational potential energy (in J) that occurs in a time of 0.20 s. (c) Find the electrical energy (in J) dissipated in the resistor in 0.20 s. Can you explain please?

Explanation / Answer

[a]

Voltage induced = Bvl=0.56*4.3*1.5= = 3.6V
Current, i = V /R = 3.6 / 0.82 = 4.40 A

Force on a conductiong rod, F= iBl= 4.40*0.56*1.5= 3.7 N

F= mg

3.7= mg

m= 3.7/ 9.8= 0.3776 kg

mass of the rod is 0.3776  kg

[b]

From the kinematic relations
s = ut+ (1/2) gt2

= 0 +(1/2)gt2
= (0.5)(9.80 m/s2)(0.20 s)2
= 0.196 m

PE lost = mgh =0.3776*9.8*0.196 = 0.725 J

[c]

power, p= vI= 3.6* 4.40= 15.84 W

E= pt= 15.84*0.20= 3.168 J

Electrical energy dissipated in the resistor in 0.20 s is 3.168 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.