Question

 

1. Two wires, each having a weight per unit length of 1.1x10^-4 N/m, are parallel with one directly above the other. The wires carry currents of 13.93 A that are equal in magnitude and opposite in direction. How far apart should the wires be placed such that the top wire will levitate above the bottom wire? Report your answer in meters.

2. Consider a -17.7 nC charge traveling next to a current-carrying wire as shown in the figure. What is the magnitude of the force (in nN) on the charge if it is moving at speed 4.49 x 10⁵ m/s at a distance of 5.67 mm from a wire carrying a 680 mA current?

3. An electron moves at 7.8x10⁶ m/s along the x-axis. It enters a region in which there is a magnetic field of magnitude 2.9 T, directed at an angle of 35 degrees with the x-axis and lying in the xy-plane. Find the magnitude of the magnetic force on the electron in pN (picoNewtons).

Answer 1

1. given:

Weight per unit length of each wire, mg/L = 1.1 x 10-4 N/m

Current through each wire, I = 13.93A

Two parallel wires carrying current experience a force, F = μo II xL/2πd

Where, L – length of the wires

                d – distance between the wires

for the upper wire just be levitated, its weight = magnetic force

                mg = F

                mg = μo II xL/2πd

                mg/L = μo II /2πd

1.1 x 10-4 = 4π x 10-7x 13.93 x 13.93/2πd

d = 4π x 10-7x 13.93 x 13.93/2πx1.1 x 10-4 = 0.358 m

answer: the distance between the two wires so that the upper wire is just levitated is

                        d = 0.358 m

2.

Given: charge, q = -17.7 nC = -17.7 x 10-9 C

Diatance of the charge from the wire, a = 5.67 mm = 0.00567 m

Speed of the charge, v = 4.49 x 105 m/s

Current in the wire, I = 680 mA = 0.68 A

The magnitude of magnetic force at a distance a from a current carrying wire,

B = μo I/2πa = 4π x 10-7x 0.68/2πx 0.0567 = 23.98 x 10-7 T

Force on a charge q moving through a magnetic field, F = qvB

      F = 17.7 x 10-9 x 4.49 x 105 x 23.98 x 10-7 = 19.05 x 10-9 N = 19.05 nN

Answer: the magnitude of the force on the moving charge, F = 19.05 nN

3.

Given:

Speed of electron, v = 7.8 x 106 m/s

Magnetic field, B = 2.9 T

Angle between magnetic field and direction of motion of charge, θ = 35 degree

Charge of electron, q = 1.6 x 10-19 C

Force on a charge q moving through a magnetic field, F = qvB Sinθ

      F = 1.6 x 10-19 x 7.8 x 106 x 2.9 x Sin35 = 2.07 x 10-12 N = 2.07 picoN

Answer: the magnitude of the force on the moving charge, F = 2.07 picoN