An electron and a proton are each moving at 765 km/sin perpendicular paths as sh

Question

An electron and a proton are each moving at 765 km/sin perpendicular paths as shown in (Figure 1) .

Part A

At the instant when they are at the positions shown, find the magnitude of the total magnetic field they produce at the origin.

Part B

Find the magnitude of the magnetic field the electron produces at the location of the proton.

Part C

Find the magnitude of the total magnetic force that the electron exerts on the proton.

Part D

Find the magnitude of the total electrical force that the electron exerts on the proton.

Part E

Find the direction of the total electrical force that the electron exerts on the proton.

Explanation / Answer

velocity of proton and electron v=765 km/sec

v=765*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*765*10^3)/(4*10^-9)^2

B1=0.765 mT

and

magnetic field due to electron is,

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

B2=(4pi*10^-7/(4pi))*(1.6*10^-19*765*10^3)/(5*10^-9)^2

B2=0.489 mT

Bnet=B1+B2

=1.2546 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*765*10^3*sin(38.66)/((4*10^-9)^2+(5*10^-9)^2)

B=0.186 mT

C)

magnetic force Fm=q*v*B

=1.6*10^-19*765*10^3*0.186*10^-3

=2.283*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

E)

and

dierctio of Fe is,

tan(theta)=5nm/4nm

===> theta=51.34 degrees,

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