All categories

2 years ago

How many turns should a 10-cm long solenoid have if it is to generate a 1.5 x 10-3 t magnetic field on 1.0 a of current?

Physics

1 answer:

stepan [7]2 years ago

3 0

We can answer this problem using Ampere’s Law:

Bh = μoNI

Where:

B = Magnetic Field

h = coil length

μo = permeability =4π*10^-7 T·m/A

N = number of turns

I = current

It is given that B=0.0015T, I=1.0A, h=10 cm = 0.1m

Use Ampere's law to find # turns:
Which can be rewritten as:
N = Bh/μoI 
N = (0.0015)(0.1)/(4π*10^-7)(1.0)
N = 119.4

Answer: 119.4 turns

You might be interested in

The moon orbits our Earth. The opposite would never be true; the Earth would never orbit the moon. One reason is due to the forc

ANEK [815]

The object with the greater mass has a greater gravitational force and that determines what satellites orbit around it. An object with more mass will never orbit an object with less.

7 0

2 years ago

Read 2 more answers

Jeff puts on a leather jacket over his sweater. The sweater becomes negatively charged. Which statements about Jeff’s situation

Vinil7 [7]

Following statements are true
(i) The leather jacket has a lower tendency to attract electrons than sweater.
When the sweater and the leather jacket are in contact with each other, the leather jacket loses electrons and thus becomes positively charged. the electrons are gained by the sweater and it becomes negatively charges.
The opposite charge attract. so the sweater ( negatively charged) will attract protons ( positively charged) . The leather jacket ( positively charged) will attract the electrons ( negatively charged).

7 0

2 years ago

Read 2 more answers

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

4 0

2 years ago

Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.

solong [7]

Answer:u=97.41m/s

Explanation:

Given

inclination $\theta =58.7^{\circ}C$

Horizontal distance travel by Particle $d=1200 m$

Vertical height $h=780 m$

Let u be the initial velocity

calculating vertical distance

$y=u\sin \theta +\frac{at^2}{2}$

$y=u\sin \theta t-\frac{gt^2}{2}$ -------1

Calculating horizontal distance

$x=u\cos \theta \times t+0$

$t=\frac{x}{u\cos \theta }$

put value of t in equation 1

$y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2$

$y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }$

$\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y$

$u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}$

$u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}$

at $y=-780\ m\ x=1200 m$

$u^2=\frac{18.154}{2753.65}\times 1200^2$

$u=97.41 m/s$

6 0

2 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

7 0

2 years ago