Practice problems for T3: 1. Find the current in the 5 0 resistors and the poten

Question

Practice problems for T3: 1. Find the current in the 5 0 resistors and the potential difference between points a and b au 52 2. A proton moving at 4.00 x 10 m/s through a magnetic field of magnitude 1.70 T experiences a magnetic force of magnitude 8.20 x 10" N. What is the angle between the proton's velocity and the field? A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00x10' m/s and experiences an acceleration of 2.00 x 10 m/s' in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field. 3. A proton (charge le, mass me), a deuteron (charge le, mass 2ms), and an alpha particle (charge 2e, mass 4m,) are accelerated from rest through a common potential difference V. Each of the particles enters a uniform magnetic field,B, with its velocity in a direction perpendicular to B. The proton moves in a circular path of radius fa. In terms of r, determine (e) R, of the circular orbit for the deuteron and (b) the radius r. for the alpha particle 4. s. A cyclotron designed to accolerate protons has an outer radius of 0 350 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the eap between the dees. The dees are between the poles of an electromagnet where the field is 0.800 T. (a) Find the cycletron frequency for the protons in this cyclotron. Find (b) the speed at which protons exit the cyclotron and (l their maximum kinetic energy

Explanation / Answer

from the given problem

1. V = 100 V

form the diagram

Reff = 4 + 10*40/(10 + 40) + 40*(5 + 5)/(40 + 5 + 5)

Reff = 20ohms

hence total curremt I = V/Reff ( from ohms law)

I = 5 A

potential difference across 2 5 ohm resistors = dV

dV = V - (4 + 10*40/50)I = 40 V

hence

coltage across 5 ohm resistor = 20 V

hence current in 5 ohm resistor, i = V/R = 20/5 = 4 A

also, potential differnce across point a and b is

va - vb = 5*i = 20 V

2. given

spee dof proton, v = 4*10^6 m/s

B = 1.7 T

F = 8.2*10^-13 N

let the angle between the protons velocity and magnetic field be theta

then

qvBsin(theta) = F

sin(theta) = 8.2*10^-13/1.6*10^-19 * 4*10^6 * 1.7

theta = 48.909856695312 deg

3. v = 10^7 m/s

theta = 90 deg

a = F/m = 2*10^13 m/s/s, in +ve x direction

v is in +z direction

now, a is in direction of v x B

hence

Bis in -y direction

magnitude

B = mp*a/qv = 1.6*10^(-27)*2*10^13/1.6*10^-19*10^7

B = 0.02 T

4. given

proton (e, m)

deuteron (e, 2m)

alpha particle (2e, 4m)

potential difference = V

so final speed = v

0.5mv^2 = qV

hence

v = sqroot(2qV/m)

hence

vp = sqroot(2eV/m)

vd = sqroot(2eV/2m) = vp/sqrt(2)

va = sqroot(2(2e)V/2(4m)) = vp/sqrt(2)

hence

form force balance in magnetic field B

mv^2/r = qvB

mv/r = qB

r = mv/qB

hence

rp = mvp/eB

rd = 2m*vp/sqrt(2)*e*B = sqrt(2)*rp

ra = 4m*vp/sqrt(2)*2e*B = sqrt(2)*rp

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