A long straight conductor carries a current I = 12.0 A running left to right. An

Question

A long straight conductor carries a current I = 12.0 A running left to right. An electron is initially located 5.00 cm from the current in the plane of the page. The electron is initially moving away from the conductor with a speed v = 175 km/s. What is the direction and magnitude of the magnetic field due to I at the initial location of the electron? What is the direction and magnitude of the initial acceleration of the electron? Now consider a proton located at the same distance of 5.00 cm from the current and initially moving parallel to the conductor in the same direction as the current. The proton is moving with an initial speed of 175 km/s. What is the initial acceleration (direction and magnitude) of the proton? What is the direction and magnitude of a uniform electric field that would allow the proton to con­tinue to move parallel to the conductor with its initial speed?

Explanation / Answer

Here,

Current , I = 12 A

d = 5 cm = 0.05 m

speed of electron , u = 1.75 *10^5 m/s

a)

magnetic field at the location = u0 * I/(2pi * d)

magnetic field at the location = 4pi *10^-7 * 12/(2pi * 0.05)

magnetic field at the location = 4.801 *10^-5 T

the direction of magnetic field is OUT of the page.

b)

for the initial acceleration of electron is a

Using second equation of motion

m * a = e * V * B

9.11 *10^-31 * a = 1.602 *10^-19 * 4.801 *10^-5 * 1.75 *10^5

a = 1.47 *10^12 m/s^2

the magnitude of initial acceleration is 1.47 *10^12 m/s^2

as F = qV X B

the direction of initial acceleartion is towards Left.

c)

for a proton

for the initial acceleration of electron is a

Using second equation of motion

m * a = e * V * B

1.67 *10^-27 * a = 1.602 *10^-19 * 4.801 *10^-5 * 1.75 *10^5

a = 8.05 *10^8 m/s^2

the magnitude of initial acceleration is 8.05 *10^8 m/s^2

as F = qV X B

the direction of initial acceleartion is towards RIght.

d)

let the electric field is E

q * E = magnetic force

1.602 *10^-19 * E = 1.602 *10^-19 * 4.801 *10^-5 * 1.75 *10^5

E = 8.401 N/C

the electric field is 8.401 N/C

the direction of electric field must be towards Left

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