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3 years ago

A solenoid has length

Physics

1 answer:

asambeis [7]3 years ago

5 0

Answer:

The magnetic field is 24.2 mT.

Explanation:

The magnetic field can be calculated with the following equation:

$B = \frac{\mu_{0}iN}{L}$

Where:

i: is the current = 5.57 A

N: is the number of turns = 5*850 turns (it is five close-packed)

L: is the length = 1.23 m

$\mu_{0}$ : is the permeability = 4π*10⁻⁷ T*m/A

Hence, the magnetic field is:

$B = \frac{4\pi \cdot 10^{-7} T*m/A*5.57 A*5*850}{1.23 m} = 0.0242 T = 24.2 mT$

Therefore, the magnetic field is 24.2 mT.

I hope it helps you!

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A student hangs a weight on a newtonmeter. The energy currently stored in the spring in the newton meter is 0.045N. The student

castortr0y [4]

Answer:

5x10^-3

Explanation:

Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.

Hooke's Law can be represented as

<em>where F is the force </em>

<em> k is the spring constant</em>

<em> x is the extension of the material </em>

<em />

Plug values in the equation

Step 1 find the original extension

0.045 = (400)x

x = 1.125x 10^-4 m d

Step 2 find the new extension

0.045+2 = 400(x)

2.045 = 400x

x = 5.1125x10^-3

Step 3 subtract the new extension with original

Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3

8 0

3 years ago

A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

And so, for r<a,

$V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

- Potential at the surface of the outer cylinder:

$V(b)=0$

Therefore, the potential difference is simply equal to

$V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

The electric field is just the derivative of the electric potential:

$E=-\frac{dV}{dr}$

so we can find it by integrating the expression for the electric potential. We find:

$E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

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#LearnwithBrainly

7 0

3 years ago

Please Help!

irga5000 [103]

The maximum force that the tires can exert on the road before slipping is 16200 N.

From the information in the question;

The coefficient of static friction = 0.9

The mass of the car = 1800 kg

Using the formula;

μ = F/R

μ = coefficient of static friction

F = force on the tires

R = the reaction force

But recall that the reaction is equal in magnitude to the weight of the car.

W=R

Hence; R = 1800 kg × 10 ms-2 = 18000 N

Making F the subject of the formula;

F = μR

Substituting values;

F = 18000 N × 0.9

F = 16200 N

Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.

Learn more: brainly.com/question/18754989

6 0

2 years ago

Diane swim 5 kilometers against the current in the same amout of time it took her to swim 15 kilometers with the current . The r

lidiya [134]

Answer:

4 km/hr

Explanation:

suppose 's' is Diane's speed with no current.

't' represents time in hrs.

Using the formula:

Distance = speed 's' x time 't'

-> when she swims against the current, equation will be,

5= (s-2)t

t= 5/(s-2)

->when she was swimming with the current, equation is,

15= (s+2) t

t= 15/(s+2)

equating eq(1) and (2)

5/(s-2) = 15/(s+2)

5s + 10 = 15s - 30

40= 10s

s= 40/10

s=4

Therefore, if there were no current, her speed is 4km/hr

5 0

3 years ago

a student pushes a 0.500 kg trolley along a frictionless surface and accelerates it from rest to 4m/s. how much kinetic energy d

aalyn [17]

mass=m=0.5kg
velocity=4m/s

$\\ \sf\longmapsto KE=\dfrac{1}{2}mv^2$

$\\ \sf\longmapsto KE=\dfrac{1}{2}(0.5)(4)^2$

$\\ \sf\longmapsto KE=0.25(16)$

$\\ \sf\longmapsto KE=4J$

7 0

2 years ago