A charged particle of mass m = 7.9X10 -8 kg, moving with constant velocity in th

Question

A charged particle of mass m = 7.9X10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 3.5T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.64 m, 0) and leaves the region at (x,y) = 0, 0.64 m a time t = 668 s after it entered the region.

1)With what speed v did the particle enter the region containing the magnetic field?

______m/s

2)What is Fx, the x-component of the force on the particle at a time t1= 222.7 s after it entered the region containing the magnetic field.

_______N

3)What is Fy, the y-component of the force on the particle at a time t1= 222.7 s after it entered the region containing the magnetic field.

_________N

4)What is q, the charge of the particle? Be sure to include the correct sign.

______C

5)If the velocity of the incident charged particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory of the particle the same?

Increase B by a factor of 2

Increase B by less than a factor of 2

Decrease B by less than a factor of 2

Decrease B by a factor of 2

There is no change that can be made to B to keep the trajectory the same.

Explanation / Answer

1) the angular velocity is constant in the movement

. w = 2 f = 2 / T

The marked region corresponds to the orbital ¼ therefore T = 4 * t = 2672 s and the radio is r = 0.64 m

Kinematics. v = r w

. v /r = 2 /T v = 2 r/T = 1.5 10-3 m/s

(4) we estimate charge using magnetic force in newton's second law

F = q v x B = mc = m v2/r

The movement of a particular charged and a magnetic field are perpendicular

. q v B Sin (90) = m v2 /r q = m v2 /r B = 7.9 10-14 C

1 c = 106 µ C

The sign is obtained from the right hand rule, where fingers are running velocity direction close in the direction of the magnetic field and thumb gives the sense of strength, this gives strength to the outside of the circle but as moving in the opposite direction the sign of q is negative.

. q = - 7.9 10-8 µ C

2 and 3) r2 = x2 + y2 r = cte = 0.64 m

. w = 2 / T = cte = 2.35 10-3 rad/s

The particule runs along an arc of (2 r) in 668 s time, if we using a rule of proportions waht recore at 222.7 s

2 r ---- 668 s

2 r´ ---- 222.7 2 r´ = 222.7 / 668 = 0.333 (2 r)

The arc is equal to 90° the fraction covered is 90° (0.333) arc = 30 °

F = Fx i + Fy j

F = q v x B = - 7.9 10-14 1.5 = 10-3 3.5 - 41.475 10-17 N

Fx = F Cos () = - 41.475 10-17 Cos (30) = - 35.92 10-17 N

Fy = F Sin() = - 20. 74 10-17 N

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