Question

An alpha particle travels at a velocity of magnitude 430 m/s through a uniform magnetic field of magnitude 0.058 T. (An alpha particle has a charge of charge of +3.2 × 10^-19 C and a mass 6.6 × 10^-27 kg) The angle between the particle's direction of motion and the magnetic field is 78°. What is the magnitude of (a) the force acting on the particle due to the field, and (b) the acceleration of the particle due to this force? (c) Does the speed of the particle increase, decrease, or remain the same?

(a)

Number

Enter your answer for part (a) in accordance to the question statement

Units

Choose the answer for part (a) from the menu in accordance to the question statement This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3m/s^3times

(b)

Number

Enter your answer for part (b) in accordance to the question statement

Units

Choose the answer for part (b) from the menu in accordance to the question statement This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3m/s^3times

(c)

Choose the answer for part (c) from the menu in accordance to the question statement remain unchanged, increase, decrease

Answer 1

Velocity v can be resolved into 2 components. vcos(in the direction of magnetic field) and vsin (perpendicular to magnetic field). Hence magnetic Lorentz force(F) acting on particle by vsin component .

F=qvBsin

where q=charge of   particle=3.2x10-19C

v=velocity of   particle=430m/s

B=magnetic field=0.058T

=780

a) force  F=qvBsin

=3.2x10-19x430x0.058xsin78

=78.06x10-19​​​​​​​N

=78.06x10-19

b) Here acceleration means centripetal acceleration because the particle describes circular motion by vsin component.

acceleration(a)=v2/r .........eq(1)

where v=velocity of  ​​​​​​​ particle=430m/s

r=radius of circle=mv/qB.............eq(2)  

where m= mass of  ​​​​​​​ particle=6.6x10-27kg

q=charge of   particle=3.2x10-19C

B=magnetic field=0.058T

r=radius of circle=mv/qB

=6.6x10-27x430/(3.2x10-19Cx0.058)

=1.52x10-4

substituting r in eq(1)

acceleration(a)=v2/r

=4302/1.52x10-4=1.22x109m/s

=1.22x109

c) By vcos component charge moves in the same direction of magnetic field with no change in velocity.(hence the overall path will be helical)