Can you guys please help with this question?

Two wires each carry 10.0 A of current (in opposite directions) and are 2.50 mm apart. What is the magnetic field 37.0 cm away a

lyudmila [28]

Answer:

see answer below

Explanation:

Before we do any kind of calculation, we need to convert the proper units of the exercise. All the units of distance must be in meters, so, let's change distance of the wire, and the magnetic field to meters:

Separation between the wires are 2.5 mm:

2.5 mm * (1 m / 1000 mm) = 0.0025 m

The distance of P from the bottom of the wires is 37 cm:

37 cm * (1 m/100 cm) = 0.37 m

The distance of P from the top of the wires is just the sum of the two distances:

R = 0.37 + 0.0025 = 0.3725 m

Now that we have the distance, we can determine the magnetic field, using the following expression:

B = B(bottom) - B(top) or just B₂ - B₁

And B = μ₀ I / 2πR

Replacing in the above expression we have:

B = μ₀ I / 2π ( 1/R₂ - 1/R₁)

Now we can determine the magnetic field:

B = (4πx10⁻⁷ * 10 / 2π) (1/0.37 - 1/0.3725)

Which means that the magnetic field is out of the page.

Hope this helps

4 0

3 years ago

what is the energy of a single photon of ultraviolet light with a wavelength of 30.0 nm? 1 nm=10^-9 m

Arada [10]

From the planks equation
E=hv
V= c/ wave length
V= 3×10^8/30×10^-9
=1×10^16
E= hv
6.63×10^-34×1×10^16
= 6.63×10^-18

4 0

3 years ago

If the mass of the moon is ...

bulgar [2K]

C decreased the factor cuz the max is smaller

8 0

3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago

Find the pressure if a force of 2N is applied to an area of 0.00004m^2

BigorU [14]

Answer:

5 x 10^4 N/m^2

Explanation:

Pressure, force and area are related witg the following equation;

Pressure = Force /Area

From the question, we obtained the following information;

Force = 2N

Area = 0.00004m^2

Pressure =?

Pressure = Force /Area

Pressure = 2/0.00004

Pressure = 5 x 10^4N/m^2

3 0

4 years ago