Physics, 19.03.2020 08:07 tttyson

A 2.0-m lamp cord leads from the 110-V outlet to a lamp having a 75-W lightbulb. The cord consists of two insulated parallel wires 8.0 mm apart and held together by the insulation. One wire carries the current into the bulb, and the other carries it out.

1. What is the magnitude of the magnetic field the cord produces midway between the two wires? μT

2. What is the magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie? μT

3. Compare the field between the two wires with Earth’s magnetic field (0.5×10^−4 T). Give your answer as B/BEarth.

4. Compare the field between 2.0 mm from one of the wires in the same plane in which the two wires lie with Earth’s magnetic field?

5. What magnetic force do the two wires exert on one another? μN

Answers: 2

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Answer from: julielebo8

Explanation:

l = 2 m

V = 110 V

P = 75 W

d = 8 mm = 0.008 m

i = P / V = 75 / 110 = 0.68 A

Magnetic field due to one wire

$B =\frac{\mu _{0}}{4\pi }\times \frac{2i}{d/2}$

$B =10^{-7}\times \frac{2\times 0.68}{0.004}$

B = 3.4 x 10^-5 T

Magnetic field due to second wire

B' = 3.4 x 10^-5 T

Both the magnetic field is in the same direction os the net magnetic field at the mid point is

Bnet = B + B'

Bnet = 2 x 3.4 x 10^-5

Bnet = 6.8 x 10^-5 Tesla

Magnetic filed due to one wire

$B=\frac{\mu _{0}}{4\pi }\frac{2i}{0.002}$

$B =10^{-7}\times \frac{2\times 0.68}{0.002}$

B = 6.8 x 10^-5 Tesla

Magnetic field due to the second wire

$B'=\frac{\mu _{0}}{4\pi }\frac{2i}{0.006}$

$B' =10^{-7}\times \frac{2\times 0.68}{0.006}$

B' = 2.27 x 10^-5 Tesla

The net magnetic field is given by

Bnet = (6.8 + 2.27) x 10^-5

Bnet = 9.07 x 10^-5 Tesla

B/Bearth = (6.8 x 10^-5) / (0.5 x 10^-4) = 1.36

B/Bearth = (9.07 x 10^-5) (0.5 x 10^-4) = 1.814

5. magnetic force per unit length is given by

$F=\frac{\mu _{0}}{4\pi }\frac{2i\times i}{d}$

$F =10^{-7}\times \frac{2\times 0.68\times 0.68}{0.008}$

F = 1.156 x 10^-5 N/m

Total force on 2 m length

F' = 2 x F = 2 x 1.156 x 10^-5 = 2.312 x 10^-5 N

Answer from: Quest

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.

neglecting air resistance etc etc.,

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.

vh = 8 ms-1

the vertical velocity component has an accelerating force: 'g'.

using v = u +at where a = g = 9.81 ms-2.

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.

adding vectorally gives: -

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

Answer from: Quest

No it doesnt because it can be any number.

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