7. + 1 points YF1428.P 058 My Notes Ask Your Teacher A long, straight wire carri

Question

7. + 1 points YF1428.P 058 My Notes Ask Your Teacher A long, straight wire carries a current of 2.54 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.64 cm from the wire and traveling with a speed of 8.00 x 104 m/s parallel to the wire in the direction opposite to the current, what are the magnitude and direction of the force that the magnetic field of the current exerts on the electron? (Assume that the current runs left to right and the electron is below the wire.) to the left upwards, towards the wire downwards, away from the wire to the right

Explanation / Answer

magnetic field due to a straight current carrying wire is

B = uo I/2piR

here I = 2.54 A

R is the distance of electron from Current carrying wire = 0.0464 m

so

magnetic field acting of electron = (4pi *10^-7 * 2.54)/(2* 3.14 * 0.0464)

B = 1.1 *10^-5 Tesla

Now that magnetic force F = q v B

q is charge of electron = 1.6 *10^-19 C

v is its speed = 8 *10^4 m/s

thus magnetic force F = 1.6*10^-19* 8 *10^4 * 1.1*10^-5

F = 1.4 *10^-19 N

to get the Direction of force , apply Left hand rule, upwards towards the wire.

--------------------------------------------

from that force per unit length between the parallel wires we have

F/l = uo I^2/2pid

but Force F = mg

so

mg/L = uo I^2 /(2pid)

I^2 = mg * 2pid/(uo L)

here the distance between parallel wires d is obtained from

angle = arc length/radius

length d = 5 * 0.04 = 0.2 m

I^2 = (m/L) * g * 2pid/uo

I^2 = 0.0465 * 9.81 * 2* 3.14 * 0.2/(4pi *10^-7)

I = current in each wire = 675.22 Amps

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