An electron and a proton are each moving at 825 km/s in perpendicular paths as s

Question

An electron and a proton are each moving at 825 km/s in perpendicular paths as shown in the figure (Figure 1) . At the instant they are at the positions shown in the figure.

a)Find the magnitude of the total magnetic field they produce at the origin.? mT

B) Find the magnitude of the magnetic field the electron produces at the location of the proton? T

C)Find the magnitude of the total magnetic force that the electron exerts on the proton? N

D)Find the magnitude of the total electrical force that the electron exerts on the proton.?N

E)Find the direction of the total electrical force that the electron exerts on the proton.? counterclockwise from the +xaxis

Explanation / Answer

velocity of proton and electron v=825 km/sec

v=825*10^3 m/sec

a)

at origin

magnetic field due to proton is,

B1=(uo/4pi)*(q*v)/r^2

B1=(4pi*10^-7/(4pi))*(1.6*10^-19*825*10^3)/(4*10^-9)^2

B1=0.825 mT

and

magnetic field due to electron is,

B2=(uo/4pi)*(q*v)/r^2

B2=(4pi*10^-7/(4pi))*(1.6*10^-19*825*10^3)/(5*10^-9)^2

B2=0.528 mT

Bnet=B1+B2

=1.353 mT

b)

B=(uo/4pi)*(q*v*sin(theta)/r^2

theta = tan^-(4nm/5nm)

theta = 38.66 degrees,

B=(4pi*10^-7/(4pi))*(1.6*10^-19*825*10^3*sin(38.66)/((4*10^-9)^2+(5*10^-9)^2)

B=0.201 mT

c)

magnetic force Fm=q*v*B

=1.6*10^-19*825*10^3*0.201*10^-3

=2.65*10^-17 N

D)

electrical force Fe=k*q1*q2/r^2

=9*10^9*(1.6*10^-19)^2/((4*10^-9)^2+(5*10^-9)^2)

=5.62*10^-12 N

and

dierctio of Fe is,

tan(theta)=5nm/4nm

===> theta=51.34 degrees,

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