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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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kenny6666 [7]

Answer:

induced emf =  28.65 mV

Explanation:

given data

diameter = 7.3 cm

magnetic field = 0.61

time period = 0.13 s

to find out

magnitude of the induced emf

solution

we know radius is diameter / 2

radius = 7.3 / 2

radius = 3.65 m

so induced emf is dπ/dt  = Adb/dt

induced emf =  A × ΔB / Δt

induced emf =  πr² × ΔB / Δt

induced emf =  π (0..65)² × ( 0.61 - (-0.28))  / 0.13

induced emf =  0.0286538 V

so induced emf =  28.65 mV

3 0
3 years ago
Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal
Harrizon [31]

Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.

b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

8 0
3 years ago
Kinetic Energy - What does it depend on?
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Answer:

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Explanation:

7 0
3 years ago
a 3000 kg and a 7000 kg Mass attract each other with a force of 0.0015 N. What distance separates the two objects (Radius) (Plea
frutty [35]

Answer:

<h3> 3.057m</h3>

Explanation:

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d² = 6.67408 × 10-11 *3000*7000/0.0015

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d² = 1.4016*10^-3/0.0015

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-- I bet it'll fit now.

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8 0
3 years ago
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