Magnesium and sulfur Ionic or covalent?

A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie

Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $$\theta$

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $$\theta$

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $$\theta$

U = Iπ $r^{2}$ B cos $$\theta$

For $$\theta$ = 0° →

U(Max) = MB cos(0) = MB = Iπ $r^{2}$ B = 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

375 π x $10^{-7}$ .

For $$\theta$ = 90° →

U = MB cos (90) = 0

For $$\theta$ = 180° →

U(Min) = MB cos(0) = - MB = - Iπ $r^{2}$ B = - 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

- 375 π x $10^{-7}$ .

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

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3 0

1 year ago

You can obtain a rough estimate of the size of a molecule by the following simple experiment. Let a droplet of oil spread out on

Marat540 [252]

Answer:

The diameter of the molecule of oil is $1.405 10 ^{-9} \ m$

Explanation:

We define density as

$\rho = \frac{mass}{volume}$

So, the volume for our oil will be

$volume = \frac{mass}{\rho}$

$volume = \frac{7.62 \ 10^{-7} \ kg}{ 912 \ \frac{kg}{m^3} }$

$volume = 8.355 \ 10 ^{-10} \ m^3$

the volume for a cylinder with radius r and height h is

$volume = \pi r^2 h$

So, we can obtain the height of the droplet of oil as:

$h = \frac{volume}{\pi r^2}$

the radius is

$r=43.5 \ cm = 0.435 \ m$

$h = 1.405 10 ^{-9} \ m^3$

And this is the diameter of the oil molecule.

6 0

3 years ago

If a proton were released from rest at the sphere's surface, what would be its speed far from the sphere?

balu736 [363]

Let the sphere is having charge Q and radius R

Now if the proton is released from rest

By energy conservation we can say

$U = K$

$\frac{kQe}{R} = \frac{1}{2}mv^2$

$\frac{2kQe}{mR} = v^2$

now take square root of both sides

$v =\sqrt{\frac{2kQe}{mR}}$

so the proton will move by above speed and

here Q = charge on the sphere

R = radius of sphere

$k = 9 * 10^9$

5 0

2 years ago

Read 2 more answers

An egg drops from a nest at a height of 2.3 m. Which equation can you use to calculate the impact speed of the egg?

igomit [66]

Answer:

V2^2=V1^2+2a/\d

Explanation:

3 0

3 years ago

A 55.0-g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum? Express

andrew11 [14]

Answer:

Final temperature of the aluminum = 41.8 °C

Explanation:

We have the equation for energy

E = mcΔT

Here m = 55 g = 0.055 kg

ΔT = T - 27.5

Specific heat capacity of aluminum = 921.096 J/kg.K

E = 725 J

Substituting

E = mcΔT

725 = 0.055 x 921.096 x (T - 27.5)

T - 27.5 = 14.31

T = 41.81 ° C = 41.8 °C

Final temperature of the aluminum = 41.8 °C

6 0

3 years ago