show me step by step please A smooth, frictionless track is placed on the ground

Question

show me step by step please

A smooth, frictionless track is placed on the ground. A uniform magnetic field having a direction shown in the picture is applied everywhere. The strength of the magnetic field is B = 2 Tesla. The height of the track is h = 96 m. A particle with mass m = 3 kg and charge q = +0.2 C starts to slide down the track from the top at point a. The initial velocity of this particle is zero. You can ignore all friction. When the particle reaches the bottom of the track at point b, what is the velocity v of this particle? At the bottom b, how much is the Lorentz force Fb on the particle? At the bottom b, how much is the normal force N from the ground on the particle? If the track is very high, the particle will gain a large velocity at the bottom b so that the Lorentz force might be bigger than the weight of the particle. Therefore, the Lorentz force would lift the particle up in the air. There should be a critical height ho such that when the particle reaches the bottom b it is about to be lifted up but remains in good contact with the ground. In other words, if you choose ho right, the normal force from the ground at point b (and thereafter) is zero. What is this critical height ho of the track?

Explanation / Answer

a) magnetic field doesnt do any work so

KE = m g h

1/2 mv^2 = m g h

v = sqrt(2 g h)= sqrt(2*9.81*96)=43.4 m/s

b) F = q v B = 0.2*43.4*2=17.36

c) sum forces in the y

N + F magnetic - mg = 0

N = 3*9.81-17.36 =12.07 N

d)

if N = 0, F magnetic = mg

q v b = m g

v = mg/( q B) = 3*9.81/(0.2*2)=73.58 m/s

and we know that 1/2 mv^2 = m g h

h = v^2/(2 g) = 73.58^2/(2*9.81)=275.9 m

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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