Help me anybody please solve this

A 8.0-cm long solenoid has 2000 turns of wire and carries a current of 5.0-A. Calculate the strength of the magnetic field at th

CaHeK987 [17]

Answer:

B = 0.157 T

Explanation:

Given that,

Length of the solenoid, l = 8 cm = 0.08 m

Number of turns in the wire, N = 2000

Current, I = 5 A

We need to find the strength of the magnetic field at the center of the solenoid. It is given by the formula as follows :

$N=\mu_o nI$ , N is number of turns per unit length of solenoid.

So,

$B=4\pi \times 10^{-7}\times \dfrac{2000}{0.08}\times 5\\\\=0.157\ T$

So, the magnetic field at the center of the solenoid is 0.157 T.

6 0

3 years ago

If the pH of a solution decrease's, is the solution getting more acidic or lees acidic

enyata [817]

Answer:

The solution becomes more acidic.

Explanation:

As the pH of a solution decreases, the concentration of hydrogen ions [H+] increases. Acidic solutions have a higher concentration of hydrogen ions and a lower pH.

I hope this helped. :)

6 0

3 years ago

Which most likely appear in the central nervous system only?

BabaBlast [244]

Answer:

Interneurons

Explanation:

8 0

4 years ago

Read 2 more answers

Which particle is used as a beam to treat cancer?

larisa [96]

Proton beams are what is used to treat cancer

5 0

3 years ago

Read 2 more answers

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words: