An electron accelerated from rest through a voltage of 400 V enters a region of

Question

An electron accelerated from rest through a voltage of 400 V enters a region of constant magnetic field.

#1

If the electron follows a circular path with a radius of 16 cm , what is the magnitude of the magnetic field?

Express your answer using two significant figures.

A proton with a kinetic energy of 5.0×1016 Jmoves perpendicular to a magnetic field of 0.26 T .

#2

What is the radius of its circular path?

Express your answer using two significant figures.

An electron accelerated from rest through a voltage of 400 V enters a region of constant magnetic field.

#1

If the electron follows a circular path with a radius of 16 cm , what is the magnitude of the magnetic field?

Express your answer using two significant figures.

A proton with a kinetic energy of 5.0×1016 Jmoves perpendicular to a magnetic field of 0.26 T .

#2

What is the radius of its circular path?

Express your answer using two significant figures.

Explanation / Answer

Q1.

magnitude of kinetic energy of the electron=q*V

where q=magnitude of charge on electron=1.6*10^(-19) C

V=voltage applied=400 volts

mass of electron=m=9.1*10^(-31) kg

then if speed acquired before entering the magnetic field region is v m/s,

then 0.5*m*v^2=q*V

==>v=sqrt(2*q*V/m)

inside the magnetic field, magnetic force is balanced by the centripetal force

radius of the path=r=0.16 m

balancing the forces,

q*v*B=m*v^2/r

where B=strength of the magnetic field

==>B=m*v/(q*r)=m*sqrt(2*q*V/m)/(q*r)

using the values,

B=9.1*10^(-31)*sqrt(2*1.6*10^(-19)*400/(9.1*10^(-31)))/(1.6*10^(-19)*0.16)

=4.2159*10^(-4) T

in two significant figures,

answer=4.2*10^(-4) T

Q2.mass of the proton=1.67*10^(-27) kg

let speed of the proton=v m/s

kinetic energy=E=5*10^(-16) J

magnetic field=B=0.26 T

charge=q=1.6*10^(-19) C

let radius of the path=r m

kinetic energy=E

==>0.5*m*v^2=E

==>v=sqrt(2*E/m)

balancing magnetic and centripetal force

q*v*B=m*v^2/r

==>r=m*v/(q*B)

=m*sqrt(2*E/m)/(q*B)

=1.67*10^(-27)*sqrt(2*5*10^(-16)/(1.67*10^(-27)))/(1.6*10^(-19)*0.26)

=0.031065 m

in two significant figures,

answer=3.1*10^(-2) m

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