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2 years ago

7

A long solenoid that has 1000 turns uniformly distributed over a length of 0.400 m produces a magnetic field of magnitude 1.00 ×

10⁻⁴T at its center. What current is required in the windings for that to occur?

1 answer:

yKpoI14uk [10]2 years ago

7 0

0.03185A current is required in the windings for that to occur.

Solenoids are essentially coils of cord. These generate a magnetic subject which strives a pressure over a steel element.A solenoid is a fundamental time period for a coil of cord that we use as an electromagnet.
We additionally consult with the tool which could convert electric power into mechanical power as a solenoid.

Actually it generates a magnetic subject for developing linear movement from the electrical cutting-edge. With using a magnetic subject.

Magnetic field at the centre of a solenoid of length 'L' having 'N' number of turns with a current 'I' is given by $B=\frac{u_0NI}{L}$ ...(1)

It is given that 1000 turns uniformly distributed over a length of 0.400 m produces a magnetic field of magnitude 1.00 ×10⁻⁴T .

Putting above values in equation (1) , we get

$1.00\times10^{-4}=\frac{4\pi \times10^{-7}\times1000I}{0.400} \\\\I=\frac{1.00\times10^{-4} \times0.400}{4\pi \times1000} \\\\I=0.03185A$

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The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

<u>Given the following data:</u>

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

<u>Conversion:</u>

1 day = 24 hours

24 hours to minutes = $60$ × $24$ = $1440$ minutes

$1440 + 37 = 1477 \; minutes$

1 minute = 60 seconds

1477 minute = X seconds

Cross-multiplying, we have:

$X = 60$ × $1477$

X = 88620 seconds

Now, we can find the frequency of rotation of Mars by using the formula:

$Frequency = \frac{1}{Period}\\\\Frequency = \frac{1}{88620}$

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Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.

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3 years ago

When popcorn is heated over a flame, the kernel burst open. why does this occur?

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3 years ago

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2 years ago

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

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3 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago