Ask question

Ask question

All categories

aleksklad [387]

2 years ago

14

What current is required in the windings of a long solenoid that has 1580 turns uniformly distributed over a length of 0.44 m in

order to produce a magnetic field of magnitude 0.000394 T at the center of the solenoid

1 answer:

Dominik [7]2 years ago

6 0

The current required in the windings of the given long solenoid that has 1580 turns uniformly distributed over the length of the solenoid is 1.98 A.

<h3>Magnetic field at the center of a solenoid</h3>

The magnetic field at the center of a solenoid is given by the following formula,

B = μ₀nI

where;

μ₀ is permeability of free space
n is number of turns
I is current in the solenoid

<h3>Current in the solenoid</h3>

I = B/μ₀n

I = (0.00394) / (4π x 10⁻⁷ x 1580)

I = 1.98 A

Thus, the current required in the windings of the given long solenoid is 1.98 A.

Learn more about current in solenoid here: brainly.com/question/17086348

You might be interested in

Draw a neat labelled diagram for a particle moving in a circular path with a constant speed. In your diagram show the direction

Naya [18.7K]

Explanation:

Particle moving in a circular path with a constant speed.

3 0

3 years ago

Which phrase best describes a machine?

Nastasia [14]

Answer:

All of the above.

Explanation:

All Machines make work easier.

They can move objects.

They have multiple parts.

And the use energy.

5 0

3 years ago

Read 2 more answers

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for l

liberstina [14]

Explanation:

Given: $\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}$

$\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}$

$\:\:\:\:\:= 6×10^{14}\:\text{Hz}$

The work function $\phi$ is then

$\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})$

$\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}$

3 0

2 years ago

valkas [14]

I try to aim a the right key or do typing lessons

5 0

3 years ago

Read 2 more answers