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A particle with negative charge q and mass m = 2.58 * 10-15 kg is traveling through a region containing a uniform magnetic fieldB→=−(0.120T)k. At a particular instant of time the velocity of the particle isv→=(1.05×106m/s)(−3i^+4j^+12k^ and the force on the particle has a magnitude of 2.45 N. (a) Determine the charge q. (b) Determine the acceleration of the particle. (c) Explain why the path of the particle is a helix, and determine the radius of curvature R of the circular component of the helical path. (d) Determine the cyclotron frequency of the particle. (e) Although helical motion is not periodic in the full sense of the word, the x- and y-coordinates do vary in a periodic way. If the coordinates of the particle at t = 0 aredetermine its coordinates at a time t = 2T, where T is the period of the motion in the xy-plane.

Short Answer

Expert verified
  1. The charge is-3.89×10-6C
  2. The acceleration of the particle is(7.60×1014m/s2)i^+(5.70×1014m/s2)j^
  3. The radius of the curvature is 0.0290m
  4. The cyclotron frequency of the particle is2.88×107Hz
  5. The coordinates of the particle is (x,y,z) = (R,0,0.874 m)

Step by step solution

01

Force acting on a particle

The force acting on a particle is given by

F→=qv×B→

02

Determine the charge of the particle

(a)

The force acting on the particle is

F→=qv→×B→=q(−0.120T)(1.05×106m/s)(−3i^×k^+4j^×k^+12k^×k^)=−q(1.26×105N/C)(+3j^+4i)=Fxi^++Fyj^

Magnitude of the force is

F=Fx2+Fy2=−q(1.26×105N/C)(3)2+(4)2=−5q(1.26×105N/C)

Therefore,

q=−F5(1.26×105N/C)=−2.45N5(1.26×105N/C)=−3.89×10−6C

03

Determine the acceleration of the particle

(b)

The force acting on the particle is

F→=ma→

Or

a→=F→m

Therefore, the acceleration is

a→=F→m=−q(1.26×105N/C)(+4i^+3j^)m=−−3.89×10−6C1.26×105N/C(+4i^+3j^)m)=(0.490N2.58×10−15kg)(+4i^+3j^)=(1.90×1014m/s2)(+4i^+3j^)=(7.60×1014m/s2)i^+(5.70×1014m/s2)j^

Therefore, the acceleration of the particle is

(7.60×1014m/s2)i^+(5.70×1014m/s2)j^

04

Determine the radius of curvature

(c)

The radius of curvature is given by

F=m(v2R)R=mv2F

Where,

v2=vx2+vy2=(−3.15×106m/s)2+(+4.20×106m/s)2=2.756×1013m2/s2AndF=Fx2+Fy2=(0.490N)42+32=2.45NThereforeR=mv2F=(2.58×10−15kg)(2.756×1013m2/s2)2.45N=0.0290m

Therefore, the radius of the curvature is 0.0290 m

05

Determine the cyclotron frequency

(d)

The cyclotron frequency is given by

f=v2Ï€¸é=vx2+vy22Ï€¸é=5.25×106m/s2Ï€(0.0290m)=2.88×107Hz

Therefore, the cyclotron frequency of the particle is2.88×107Hz

06

Determine the coordinates of the particle

(e)

The time duration is

T=1f=12.88×107Hz=3.47×10−8s

The coordinates will be

z=z0+vzt=0+(12.6×106m/s)(2)(3.47×10−8s)=0.874m

Therefore, the coordinates of the particle is(x,y,z)=(R,0,0.874m)

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