An electron and a proton are each moving at 895 km/s in perpendicular paths as s

Question

An electron and a proton are each moving at 895 km/s in perpendicular paths as shown in the following figure. Consider the instant when they are at the positions shown in the figure. (Let dx = 3.70 nm and dy = 5.45 nm.)

(a) Find the magnitude and direction of the total magnetic field they produce at the origin.

(b) Find the magnitude and direction of the magnetic field the electron produces at the location of the proton.

(c) Find the magnitude and direction of the total electrical force and the total magnetic force that the electron exerts on the proton.

Explanation / Answer

given data

velocity v=895 km/s

dx = 3.70 nm

dy = 5.45 nm.

q =1.6*10-19 C

a)

the magnitude and direction of the total magnetic field at the origin.

Bx+By= (0/4)qv*(1/dy2 +1/dx2)

=10^-7(1.6*10^-19*895)(1/5.45^2+1/3.7^2)

   =1.52mT

in to the page

b)

theta=tan inverse(dy/dx)=55.8

magnetic field

B =(0/4)*qvsin theta/(dx2+ dy2),

=10^-7*1.6*10^-19*895*sin 55.8/(3.7^2+5.45^2)

=2.72*10^-4 T

into the page

c)

magnetic force Fm=qv*B

=4.11*10^-17 N

direction is +x

eletric field

E= kq2/(dx2 +dy2)

=6.01*10^-12 N/C

electric force

Fe=q*E=10.15*10^-31 N

direction

360-60=300 degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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