Motion In a Magnetic Field 1 A charged particle of mass m- 5.3x10 kg, moving wit

Question

Motion In a Magnetic Field 1 A charged particle of mass m- 5.3x10 kg, moving with constant velocity in the y direction enters a region containing a constant magnetic field B-2.5T aligned with the positive z-axis as shown. The particle enters the region at y) (0.56 m, 0) and leaves the region at (x,y) . 0, 0.56 m a time t . 763 ?s after?t entered the (0 ,(00 )With what speed v did the particle enter the region containing the magnetic field? m/s Submit 2) What is Fu, the x -component of the force on the particle at a time ty -254.3 us after it entered the region containing the magnetic field. Tipler6 a) what is Fy, the y component of the force on the particle at a time ty 2543 js after it entered the region containing the magnetic field 4) What is q, the charge of the particle? Be sure to include t the correct sign MacBook Air 2.0 888 ,a 2 3 4 5 6 9

Explanation / Answer

For the given data

m = 5.3*10^-8 kg

its moving in +y direction with speed v

B = 2.5 T ( along + ve z axis)

the location of entering the magnetic field (0.56m, 0)

the location of exiting the magnetic field (0,0.56 m)

t = 763 us

1. hence

v = pi*R/2*t

R = 0.56 m

hence

v = 1152.8780380146 m/s

2. at t1 = 254.3 us

angle with the horizontal of the position vector = phi

hence

theta = w*t = v*t/r = 1152.8780*763*10^-6/0.56 = 0.523530151 rad

hence

x component of force = -qvB*cos(theta)

now, qvB = mv^2/r

qvB = mv^2/r

hence

Fx = -mv^2*cos(theta)/r = -0.108943773026 N

3. the Y component is Fy = -mv^2*sin(theta)/r = -0.06288874878116 N'

4. q = mv/Br

the charge is -ve from the direciton of the force

hence

q = -4.36*10^-5 C

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

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