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tino4ka555 [31]
2 years ago
14

1. Magnetic force acting on a unit positive charge moving perpendicular to the magnetic field with a unit velocity is called

Physics
1 answer:
OlgaM077 [116]2 years ago
6 0

Magnetic Flux is the magnetic force acting on a unit positive charge moving perpendicular to the magnetic field with a unit velocity

<h3>What does magnetic flux depend on ?</h3>

The magnetic flux or field lines linking a coil depends on;

  • The magnetic field strength
  • The number of turns of the coil
  • The area of each turn

According to the question, the Magnetic force acting on a unit positive charge moving perpendicular to the magnetic field with a unit velocity is called magnetic flux

To obtain a large induced electromotive force, we cause a coil of many turns to move at a high speed across a strong magnetic field.

According to Faraday's law, which state that whenever there is a change in the magnetic lines of force, an electromotive force is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit.

Therefore, Magnetic Flux is the magnetic force acting on a unit positive charge moving perpendicular to the magnetic field with a unit velocity

Learn more about Magnetic Flux here: brainly.com/question/16234377

#SPJ1

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One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to t
WINSTONCH [101]

Complete question is;

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?

Answer:

B_net = 50 × 10^(-7) T

Explanation:

We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).

This means that the second wire is 4 m in length on the positive y-axis.

Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.

This means that the position we want to find is half the length of the second wire.

Thus, at this point the net magnetic field is given by;

B_net = √[(B1)² + (B2)²]

Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.

Now, formula for magnetic field due to very long wire is;

B = (μ_o•I)/(2πR)

Thus;

B1 = (μ_o•I_1)/(2πR_1)

Also, B2 = (μ_o•I_2)/(2πR_2)

Now, putting the equation of B1 and B2 into the B_net equation, we have;

B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]

Now, factorizing out some common terms, we have;

B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]

Now,

μ_o is a constant and has a value of 4π × 10^(−7) H/m

I_1 = 30 A

I_2 = 40 A

Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.

Thus;

R_1 = 2 m

R_2 = 2 m

So, let's calculate B_net.

B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]

B_net = 50 × 10^(-7) T

5 0
3 years ago
Parallel conducting tracks, separated by 1.50 cm, run north and south. There is a uniform magnetic field of 1.20 T pointing upwa
Y_Kistochka [10]
I WISH I CAN HELP IM SO
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5 0
3 years ago
A car travels along a straight road, heading east for 1 h, then traveling for 30 min on another road that leads northeast. If th
Bogdan [553]

Answer:

The car is 72.75 miles away from its starting position.

Explanation:

First, remember the relation:

distance = time*speed.

Also, the distance between two points (a, b) and (c, d) is:

D = √( (a - c)^2 + (b - d)^2)

For this problem, we can assume:

The North is equivalent to the y-axis, and the East is equivalent to the x-axis.

We also assume that the initial position of the car is (0mi, 0mi)

Now the car moves to the East at a speed of 52mi/h for one hour, then the new position of the car is:

(0mi, 0mi) + (52mi/h*1h, 0mi) = (52mi, 0mi)

Now the car travels 30 mins (or 0.5 hours) to the northeast at a speed of 52mi/h.

We can assume that it moves at an exact angle of 45° from East to North, then the components of the speed can be written as:

Sx = speed in the x-axis = 52mi/h*cos(45°) = 36.77 mi/h

Sy = speed in the y-axis = 52mi/h*sin(45°) =  36.77 mi/h

Then the new position of the car is:

(52mi, 0mi) + (36.77 mi/h*0.5h, 36.77 mi/h*0.5h) = (70.385 mi, 18.385 mi)

Now we know the final position of the car.

The distance between the final position (70.385 mi, 18.385 mi) and the initial position (0mi, 0mi) is:

D = √( (70.385 mi - 0mi)^2 + (18.385 mi - 0mi)^2) = 72.75 mi

The car is 72.75 miles away from its starting position.

7 0
3 years ago
I need HELP Please !!
aliya0001 [1]
You have to figure it out
5 0
3 years ago
Suppose you apply a force of 75 N to a 25-kg object. What will the acceleration of the object be?
Serjik [45]

Answer:

I belive it would be ture

Explanation:

It's been a while since I learned this but I think that is right.

6 0
3 years ago
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