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Dafna11 [192]
1 year ago
8

Two long parallel wires carry currents of 20 a and 5. 0 a in opposite directions. the wires are separated by 0. 20 m. what is th

e magnetic field midway between the two wires? group of answer choices
Physics
1 answer:
kogti [31]1 year ago
3 0

The magnetic field will be given by the formula B=B1+B2.

Between two wires, the magnetic field has the formula B=B1+B2.

We must first calculate the magnetic field caused by each wire before adding the two vectors to find the net magnetic field halfway between the two wires. We must subtract the two vectors in order to find the total magnetic field because the electric current flows in the opposite direction via both cables.

The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is traveling through a magnetic field.

To learn more about the magnetic field please visit -
brainly.com/question/23096032
#SPJ4

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Assume you have an ice cube and also a small rock that is the same size and shape as the ice cube. Predict what would happen if
Step2247 [10]

Answer: The ice cube would float on top of the water and the rock would sink to the bottom.

Explanation: The ice cube has a smaller density than the rock which allows the ice cube to float but makes the rock sink to the bottom of the glass of water.

7 0
3 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
3 years ago
A spaceship far from all other objects uses its impulse power system to attain a speed of 104 m/s. The crew then shuts off the p
asambeis [7]

Answer:

Velocity remains the same at 104 m/s

Explanation:

According to Newton's 1st law of motion, an object subjected to no force or net force equal 0 would maintain its velocity. In our case the crew shuts off the power, spaceship is in space and far from all other objects (so no gravity whatsoever) would have no force acting on it. Therefore its velocity would stay the same at 104 m/s

4 0
2 years ago
In Ampere's law, ∮????⃗∙????????⃗=????0???? the direction of the integration around the path:
kupik [55]

Answer:

C) must be such as to follow the magnetic field lines.

Explanation:

Ampere's circuital law helps us to calculate magnetic field due to a current carrying conductor. Magnetic field due to a current forms closed loop around the current . If a  net current of value I creates a magnetic field B around it , the line integral of magnetic field around a closed path becomes equal to μ₀ times the net current . It is Ampere's circuital law . There may be more than one current passing through the area enclosed by closed curve . In that case we will take net current by adding or subtracting them according to their direction.

It is expressed as follows

∫ B.dl = μ₀ I . Here integration is carried over closed path . It may not be circular in shape. The limit of this integration must follow magnetic field lines.

the term ∫ B.dl is called line integral of magnetic field.

3 0
3 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
vodka [1.7K]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

8 0
3 years ago
Read 2 more answers
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