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Sonbull [250]
3 years ago
5

1. Consider the point exactly halfway between the two wires. Can you adjust the current so that, with current passing through ea

ch of the wires, the net field is zero here? 2. Note the distance between the field lines from one of the wires. Are they equally space? Explain. 3. With currents of different magnitudes passing through the two wires, how do the forces on the two wires compare? What is the reason for this?

Physics
1 answer:
chubhunter [2.5K]3 years ago
8 0

Answer:

hello your question is incomplete attached below is missing part of the question

answer:

1 )  Magnetic field due to long current carrying wire : B = \frac{U_{0} I}{2\pi d}

Therefore the net magnetic field due the both wires ; B = B_{1} + B_{2} . when we adjust the current I_{1} = I_{2} then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

F = \frac{U_{0}I_{1}I_{2}   }{2\pi d }

Explanation:

1 )  Magnetic field due to long current carrying wire : B = \frac{U_{0} I}{2\pi d}

Therefore the net magnetic field due the both wires ; B = B_{1} + B_{2} . when we adjust the current I_{1} = I_{2} then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

F = \frac{U_{0}I_{1}I_{2}   }{2\pi d }

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A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glas
Eduardwww [97]

Answer:

The time of short pulse of light beam is 2.37\times10^{-9}\ sec

Explanation:

Given that,

A beam of light from a laser illuminates a glass.

Suppose, the length of piece is L=25.21\times10^{-2}\ m

Index of refraction is 2.83.

We need to calculate the speed of light pulse in glass

Using formula of speed

v=\dfrac{c}{\mu}

Put the value into the formula

v=\dfrac{3\times10^{8}}{2.83}

v=1.06\times10^{8}\ m/s

We need to calculate the time of short pulse of light beam

Using formula of velocity

v=\dfrac{d}{t}

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}

t=2.37\times10^{-9}\ sec

Hence, The time of short pulse of light beam is 2.37\times10^{-9}\ sec

3 0
4 years ago
Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black
arlik [135]

Answer:

833.4801043*10^6N on ear that is closer to Black hole.

13.83803929*10^6N On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

                                                 F =\frac{m_{1}*m_{2}  }{r^2} G

                    m_{1}= 5*1.989*10^30kg is mass of Black hole.

                     m_{2}= 0.030kg is mass of ear that is close to black hole.

                    G = 6.679*10^-11m^3*kg^-1*s^-2

                     r = 120km-\frac{6}{100000} km=119.99994km=119999.94m

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

                                      F = 833.4801043*10^6N

       That is tension of ear.

(2) Force on ear that is further from black hole.

                              F =\frac{m_{1}*m_{2}  }{r^2} G

                    m_{1}= 5*1.989*10^30kg is mass of Black hole.

                     m_{2}= 0.030kg is mass of ear that is close to black hole.

                    G = 6.679*10^-11m^3*kg^-1*s^-2

         this time r is further away from black hole so it would be.

                       r = 120km+\frac{6}{100000} km = 120000.06m

Plugging this all in we get

                          F = 13.83803929N

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

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