The figure below shows a wooden cylinder with mass m = 0.225 kg and length L = 0

Question

The figure below shows a wooden cylinder with mass m = 0.225 kg and length L = 0.300 m, with N = 50.0 turns of wire wrapped around it longitudinally, so that the plane of the wire coil contains the long central axis of the cylinder. The cylinder is released on a plane inclined at an angle Q to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude 0.750 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane? A

Explanation / Answer

given data

m=0.225kg
L=0.3m
N=50
B=0.750T
I=?

if you draw free body diagram, weight of the object will act down ward and resolve it...you will be getting

2NiLB sin = mgcos

mgsin = mg cos
2NiLB=mg

2*50*i*0.3*0.750=0.225*9.8

i=0.098A

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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