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aleksklad [387]

1 year 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]1 year 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

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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

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6 0

1 year ago

Assume a nation's nominal GDP is $5 trillion with an inflation rate of 5%. find its real GDP. PLEASE HELPPPPPPPPPPPPPPPPPPPPP

erastova [34]

Answer:

I don't know

Explanation:

But if you need help or reference America's GDP is 19.8 trillion dollars it spends over 2 to 3 percent of its budget on the military and the military's budget is 250 billion dollars

PS: sorry if the reference didn't help and if this doesn't help write dow the problem and download PHOTO MATH take a picture and buala and if you need to show your work just click on the button that says brake down problem and it'll show you and teach you the steps

7 0

2 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

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2 years ago

A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer the answer to the neares

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Answer:

speed = 7.9 m/s

Explanation:

speed = total distance / time taken

speed = 300 / 38

speed = 7.89473684 m/s

to the nearest tenth

speed = 7.9 m/s

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