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Solution available with explanation
Typology: Exercises
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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
