All categories

2 years ago

When an electric current flows through a wire a

Physics

1 answer:

Rama09 [41]2 years ago

3 0

Answer:

Option C. is correct

Explanation:

The magnetic field is the area around a magnet in which there is magnetic force. When an electric current flows through a wire a magnetic field is created. A single wire does not produce a strong magnetic field. So, to increase the power of the magnetic field, increase the number of coils in the wire.

You might be interested in

Assuming that the circuit is in the steady state, what is the total energy stored in the two capacitors? j

Neporo4naja [7]

<span>The total energy stored is the sum of the energy stored in the capacitors. If the capacitors are series connected capacitors, then the charging current is the same for both capacitors. This means that each capacitor stores the same energy and the stored energy is two times the energy of any of the capacitors.</span>

8 0

3 years ago

Fernanda is working on a physics project. she is experimenting by rolling a marble down a ramp and then seeing how far it rolls

Ratling [72]

The distance covered on the floor after leaving the ramp is the dependent variable.

As a result of the marble's size, the substance it is constructed of, and the angle at which it is placed onto the ground, the distance it rolls varies.
Therefore, the angle at which the marble is released onto the ground, the type of material used to make the stone, or its size can all be considered independent variables.

<h3>What is Independent variable?</h3>

There are independent and dependent variables in every experiment.
A variable is considered independent if its change is not influenced by the change in another variable or factor.

<h3>What is Dependent variable?</h3>

In any experiment, the dependent variable must be measured or determined, and it must change as the independent variable does.

Learn more about independent and dependent variable here:

brainly.com/question/1479694

#SPJ4

5 0

1 year ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

2 years ago

When falling through the air, which of the following objects will hit the ground first?

Pavel [41]

Answer:

a penny

Explanation:

5 0

3 years ago

Read 2 more answers

A frog jumps for 4.0 seconds at a maximum horizontal distance of 0.8m. what is its velocity along the road?

Lerok [7]

Answer:

The frog's horizontal velocity is 0.2 m/s.

Explanation:

To solve this problem, we must first remember what velocity is and how we solve for it. Velocity can be solved for using the formula x/t, where x represents horizontal distance and t represents time (in seconds), that it takes to travel this distance. If we plug in the given numbers for these variables and solve, we get the following:

v = x/t

v = 0.8m/4s

v = 0.2 m/s

Therefore, the correct answer is 0.2 m/s. We can verify that these units are correct because the formula calls for distance divided by time, so meters per second is a sensible answer.

Hope this helps!

3 0

2 years ago