Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Solution available with explanation
Typology: Exercises
1 / 5
This page cannot be seen from the preview
Don't miss anything!
Click Here
Given that -
is current in wire with opposite direction.
mass of electron.
charge on electron.
v is velocity of electron.
i 1 and i 2 = 6 A
i 1 and i 2
me = (9.1 × 10−31) Kg
me
qe = (1.6 × 10−19) C
qe
v = 10^6 (
m s
wire 1 wire 2
Magnetic flux density / Magnetic field due to a straight wire is -
B is the magnetic flied. r is the distance from point P to wire. I is the current in wire.
is the constant and permeability of free space. For wire 1 magnetic field at point P is -
direction : inside the page ( exam sheet ). For wire 2 magnetic field at point P is -
direction : inside the page ( exam sheet ).
Expalantion
μI 2 πr
μ = 4 π × 10−7( T m A
μ
μi 1 2 πr 1
μi 2 2 πr 2
v is the velocity of the electron.
is the charge of the electron.
is the mass of the electron.
Expalantion
So, when electron launched with velocity into P with trajectory perpendicular to the exam sheet , electron will experience a force due to the magnetic field produced by the currents in the wires.
This force will cause the electron to move in a circular path with radius is -
is the centripetal force. m is the mass of the electron. v is the velocity of the electron. r is the radius of curvature of the path.
Force due to the magnetic field is -
is the magnetic force. q is the charge of the electron. path of electron trajectory perpendicular to the sheet so both force is equal each other for path radius.
qe = (1.6 × 10−19) C
qe
me = (9.1 × 10−31) Kg
me
mv^2 r
FM = qvB
qvB = mv^2 r
r =
mv^2 qvB
r =
mv qB
r is the radius of path so plugging value in the formula to find out the radius of path -
radius of trajectory path is 1.77m.
Magnetic flux density / Magnetic field due to two infinitely long parallel wire is -
Radius of curvature path is -
r = mv qB
r = (
) m
r = (
) m
r = (
) m
r = (
) m
r = (
) m
r = (1.77) m
r = (1.77) m
