A proton (q 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with constant velocity en

Question

A proton (q 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z- axis at (x.y)- (0,0) as shown. The magnetic field extends for a distance D 0.54 m in the x-direction. The proton leaves the field having a velocity vector (Vx. Vy)- (2.7 X 105 m/s, 1.6 X 105 m/s). 1) What is v, the magnitude of the velocity of the proton as it entered the region containing the magnetic field? 2) What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field? 3) What is h, the y co-ordinate of the proton as it leaves the region conating the magnetic field? Submit 4) What is Bz, the z-component of the magnetic field? Note that Bz is a signed number. Submit 5) If the incident velocity v were increased, how would h and e change, if at all? O h and would both increase O h and would both decrease O h would increase and would decrease O h would decrease and would increase Neither h nor would change Submit

Explanation / Answer

1)

magnitude of velocity v = sqrt(vx^2+vy^2)

v = sqrt((2.7*10^5)^2+(1.6*10^5)^2)

v = 3.14*10^5 m/s

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2)

time taken to go to point h, t = D/vx

length of the arc s = v*t = v*D/vx = 0.628 m

angle theta = tan^-1(vy/vx) = 30.6 degrees = 0.535 radians

Radius = s/theta = 0.628/0.535 = 1.17 m

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3)

along y direction

y = (voy + voy)*t/2

y = (1.6*10^5)/2*(0.54/(2.7*10^5)

y = 0.16 m

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4)

In the magnetic field

magnetic force provides necessary centripetal force

Fb = Fc

Fb = q*(v x B)

as it enters the Fbis along +y direction v is along +x direction

from right hand rule B is along -z direction

q*v*Bz = m*v^2/R

Bz = -m*v/(R*q)

Bz = -1.67*10^-27*3.14*1065/(1.6*10^-19*1.17)

Bz = -2.98*10^-5 T

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5)

h and theta would both decrease

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