Please provide full solutions. Thanks in advance! Ans: a) -5.92e-13 N in Z direc

Question

Please provide full solutions. Thanks in advance!

Ans:

a) -5.92e-13 N in Z direction

b) 4.029e14 m/s^2

c) 4.49e6 m/s

Question 8 (a) A proton is fired with an initial velocity of 3.7x10 m/s in a direction 30 degrees above the x axis. A uniform magnetic field of 2T points along the positive x direction as shown in the figure below. Find the magnitude and direction of the net force on the proton just after it has been fired. Ignore gravity 30° (b) In a separate situation, a proton is fired in the same direction as above into a region in space where, in addition to the magnetic field there is also a uniform electric field E-2x 10 V/m. Both fields point along the positive x direction. The initial velocity of the proton is 3.7x10° m/s. Find the magnitude of the acceleration of the proton. (c) Find the speed of the proton after it has travelled 17mm to the right.

Explanation / Answer

A) F = q*v*B*sin(theta)

= 1.6*10^-19*3.7*10^6*2*sin(30)

= 5.92*10^-13 N (towards -z axis)

B) here, there is no magnetic force on the proton.

only electric force acts.

Fe = q*E

m*a = q*E

ax = q*E/m

= 1.6*10^-19*2*10^6/(1.67*10^-27)

= 1.92*10^14 m/s^2

az = F/m

= 5.92*10^-13/(1.67*10^-27)

= 3.54*10^14 m/s^2

a = sqrt(ax^2 + ay^2)

= sqrt(1.92^2 + 3.54^2)*10^14

= 4.03*10^14 m/s^2

c) here magnetic force always acts perpendicular to the dispalacement.

so, workdone by magentic force = 0

Workdone by electric force = q*E*d*cos(30)

= 1.6*10^-19*2*10^6*17*10^-3*cos(30)

= 4.71*10^-15 J

use Work-energy theorem,

W = 0.5*m*(vf^2 - vi^2)

==> vf = sqrt(vi^2 + 2*w/m)

= sqrt((3.7*10^6)^2 + 2*4.71*10^-15/(1.67*10^-27))

= 4.4*10^6 m/s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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