A 5000-km long cable consists of seven copper wires, each of diameter 0.85 mm, bundled together and surrounded by an insulating
1 answer:
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Answer
given,
Electric field,E = 490 N/C
time, t = 54 ns
for electron
Mass of electron me = 9.1 x 10⁻³¹ kg
Charge of electron e = -1.6 x 10⁻¹⁹ C
electrostatic force
F = E q
F = 490 x 1.6 x 10⁻¹⁹
F = 784 x 10⁻¹⁹ N
now, using newton second law
a = 8.62 x 10¹² m/s²
using equation of motion
v = u + a t
v = 0 + 8.62 x 10¹² x 54 x 10⁻⁹
v = 4.65 x 10⁵ m/s
velocity of electron is equal to v = 4.65 x 10⁵ m/s
For Proton
Mass mp = 1.67 x 10⁻²⁷ kg
Charge p = 1.6 x 10⁻¹⁹ C
Electric field E = 490 V/C
from above solution
F = 784 x 10⁻¹⁹ N
now, acceleration
a = 4.69 x 10¹⁰ m/s²
using equation of motion
v = u + a t
v = 0 + 4.69 x 10¹⁰ x 54 x 10⁻⁹
v = 4.65 x 10³ m/s
velocity of electron is equal to v = 4.65 x 10³ m/s
Complete Question:
In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.
a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.
b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.
c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.
d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?
Answer:
force, 1.318 ₓ 10⁻⁴
direction, 18.435°
Explanation:
The attached file gives a breakdown step by step solution to the questions
