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Alex787 [66]
3 years ago
5

Two long straight wires are parallel and carry current in the same direction. The currents are 8.0 and 12 A and the wires are se

parated by 0.40 cm. The magnetic field at a point midway between the wires is:_______a. 0b. 4 x 10^-4c. 8 x 10^-4d. 12 x 10^-4e. 20 x 10^-4
Physics
1 answer:
Bess [88]3 years ago
8 0

A magnetic field is a mathematical description of the magnetic influence of electric currents and magnetic materials. The magnetic field at any point is specified by two values, the direction and the magnitude; such that it is a vector field. Mathematically it is described as,

B = \frac{\mu_0 I}{2\pi d}

Here

\mu_0=  Permeability at free space constant

I_{1,2}= Current at each object

d = Distance to the center point of the two object

Two magnetic field due to the current in the same directions then is,

B = \frac{\mu_0 I_1}{2\pi d_1}+\frac{\mu_0 I_2}{2\pi d_2}

Replacing,

B = \frac{(4\pi *10^{-7})(8)}{2\pi (2*10^{-3})}+\frac{(4\pi *10^{-7})(12)}{2\pi (2*10^{-3})}

B = 8*10^{-4} +12*10^{-4}

B = 20*10^{-4}T

Therefore the correct answer is E.

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andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

4 0
3 years ago
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Which best describes a similarity between power plants that use water as an energy source and those that use wind as an energy s
disa [49]
The hydropower plant and wind turbines both uses kinetic energy to produce mechanical power and convert the mechanical energy using a generator to an electrical energy. They both have the process to produce energy but they differ in the source the hydropower plant uses water to whit the wind turbines power plant uses wind. Therefore the answer is letter B. 
3 0
3 years ago
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I NEED HELP PLEASE, THANKS! :)
brilliants [131]

Answer:

P = VI = (IR)I = I2R

Explanation:

What the equation means is that if you double the current you end up with 4 times the power loss. It's like the area of carpet you need for a room - if you make the room twice as long and twice as wide you need 4x as much carpet. The physical explanation is that the voltage difference along a wire depends on the current - more current flowing with a resistance means more voltage (pressure of electricity if you like) is built up.

This extra voltage means more power. So if you double the current your would double the power, but you also double the voltage which doubles the power again = 4x as much power. P = VI = (IR)I = I2R

I hope this helps you out, if I'm wrong, just tell me.

8 0
3 years ago
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When two objects with electrical charges interact, which affect the strength of that interaction?
noname [10]
The two objects with electrical charges interact, which affect the strength of that interaction <span>amount of charge. The answer is letter A. The rest of the choices do not answer the question above.</span> 
6 0
3 years ago
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve
solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d =  0.107 mm =  0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

n is the electron density

so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

n = 1.54125 /  4.4396226 × 10⁻²⁶

n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

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