Ask question

Ask question

All categories

3 years ago

5

What current is required in the windings of a long solenoid that has 420 turns uniformly distributed over a length of 0.575 m in

order to produce inside the solenoid a magnetic field of magnitude 4.22×10−5 T? The permeablity of free space is 1.25664 × 10 Tm/A.

1 answer:

WARRIOR [948]3 years ago

4 0

Answer:

The current in the solenoid=0.463 Ampere

Explanation:

Given:

Number of turns , N=420

Length of the solenoid=0.575 m

Magnitude of Magnetic Field, $B=4.22\times10^{-5}\ \rm T$

We know that the magnitude of the magnetic Field inside the solenoid is

$B=\mu_0 ni$

Where

$\mu_0$ is the permeability of free space.
n is the number of turns per unit length
i is the current

According to question

$B=\mu_0 ni\\\\4.22\times10^{-5}=1.24664\times10^{-7}\times \dfrac{420}{0.575}\times i\\=0.463\ \rm Amp$

Hence the current is calculated.

You might be interested in

Please, what is a car battery pH?

Mama L [17]

Answer

about – 0.7

Explanation:

5 0

3 years ago

Lasers emit light of a certain frequency in one, precise direction. The light that a laser emits can be tuned to have a high fre

makkiz [27]

Answer:

An ultra intense laser is one with which intensities greater than 1015 W cm-2 can be achieved.

Explanation:

This intensity, which was the upper limit of lasers until the invention of the Chirped Pulse Amplification, CPA technique, is the value around which nonlinear effects on the transport of radiation in materials begin to appear.

Currently, the most powerful lasers reach intensities of the order of 1021W cm-2 and powers of Petawatts, PW, in each pulse. This range of intensities has opened the door for lasers to a multitude of disciplines and scientific areas traditionally reserved for accelerators and nuclear reactors, applying as generators of high-energy electron, ion, neutron and photon beams, without the need for expensive infrastructure.

8 0

3 years ago

Read 2 more answers

An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m

Sergeeva-Olga [200]

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement 0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

= \frac{8 *0 .10}{0.40}

a = 2 m/s2

5 0

3 years ago

If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge thr

Agata [3.3K]

Answer:

Resistance in the flash tube, $R=3.97\times 10^{-3}\ \Omega$

Explanation:

It is given that,

Speed of the bullet, v = 500 m/s

Distance between one RC constant, d = 1 mm = 0.001 m

Capacitance, $C=503\ \mu F=503\times 10^{-6}\ F$

The time constant of RC circuit is given by :

$\tau=RC$

R is the resistance in the flash tube

$R=\dfrac{\tau}{C}$ ..........(1)

Speed of the bullet is given by total distance divided by total time taken as :

$v=\dfrac{d}{\tau}$

$\tau=\dfrac{d}{v}$

$\tau=\dfrac{0.001}{500}$

$\tau=0.000002\ s$

Equation (1) becomes :

$R=\dfrac{0.000002}{503\times 10^{-6}}$

$R=3.97\times 10^{-3}\ \Omega$

So, the resistance in the flash tube is $3.97\times 10^{-3}\ \Omega$ . Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

A water pipe having a 4.00 cm inside diameter carries water into the basement of a house at a speed of 1.00 m/s and a pressure o

posledela

Answer:

$8.16\ \text{m/s}$

Explanation:

$d_1$ = Initial diameter = 4 cm

$v_1$ = Initial velocity = 1 m/s

$d_2$ = Final diameter = 7.8 m

$v_2$ = Final velocity

$A$ = Area = $\pi\dfrac{d^2}{4}$

From the continuity equation we get

$A_1v_1=A_2v_2\\\Rightarrow \pi\dfrac{d_1^2}{4}v_1=\pi\dfrac{d_2^2}{4}v_2\\\Rightarrow v_2=\dfrac{d_1^2}{d_2^2}v_1\\\Rightarrow v_2=\dfrac{4^2}{1.4^2}\times 1\\\Rightarrow v_2=8.16\ \text{m/s}$

The speed of water at the second floor is $8.16\ \text{m/s}$ .

8 0

3 years ago