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kirza4 [7]
2 years ago
9

A solenoid that is 35 cm long and contains 450 circular coils 2.0 cm in diameter carries a 1.75-A current. (a) What is the magne

tic field at the center of the solenoid, 1.0 cm from the coils? (b) Suppose we now stretch out the coils to make a very long wire carrying the same current as before. What is the magnetic field 1.0 cm from the wire’s center? Is it the same as that in part (a)? Why or why not?
Physics
2 answers:
Taya2010 [7]2 years ago
6 0

Answer:

Explanation:

a )  No of turns per metre

n = 450 / .35

= 1285.71

Magnetic field inside the solenoid

B = μ₀ n I

Where I is current

B = 4π x 10⁻⁷ x 1285.71 x 1.75

= 28.26 x 10⁻⁴ T

This is the uniform magnetic field inside the solenoid.

b )

Magnetic field around a very long wire at a distance d is given by the expression

B = ( μ₀ /4π ) X 2I / d

= 10⁻⁷ x 2 x ( 1.75 / .01 )

= .35 x 10⁻⁴ T

In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .

patriot [66]2 years ago
6 0

Answer:

a) B = 2.83 mT

b) B1 = 3.5 * 10^-2 mT

Explanation:

Length of solenoid = 35 cm

= 0.35 m

Number of turns for circular coils (N) = 450

diameter = 2.0cm

= 0.02m

Radius = 0.02/2 =0.01

Current = 1.75A

a) Magnetic field is given by;

B = μoNi / l

μo = 4π * 10^-7

B = (4π*10^-7 * 450 * 1.75) / 0.35

B = (9.8960 * 10^-4) / 0.35

B = 2.83mT

The magnetic field at the center of the solenoid, 1.0cm 4th he could is 2.83mT

b) The distance from the centre if a very long wire in d.

d = 1.0cm = 0.01m

Magnetic field(B1) = μoi / 2πd

μo = 4π * 10^-7

B1 = (4π*10^-7 * 1.75) / 2π * 0.01

B1 = (2.2 * 10^-6) / 0.63

B1 = 0.3492 * 10^-4

B1 = 3.5 * 10^-5 T

B1 = 3.5 * 10^-2 mT

The two magnetic fields are not the same. This is because in the first magnetic field, the solenoid magnetic field is multiplied due to increase in the number of turns but it is not the same for the straight coil

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7 0
1 year ago
Temperature difference in the body. The surface temperature of the body is normally about 7.00 ∘C lower than the internal temper
egoroff_w [7]

Answer:

7 K.

12. 6 °F

Explanation:

Convert the individual temperatures to Kelvin (Surface temperature and internal temperature) before calculating the temperature difference of the body,

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Converting the surface and the internal temperature to temperature in Kelvin

Surface Temperature of the body (K) = (X + 273) K

Internal Temperature of the body (K) = (X + 7) + 273 = (X + 280) K.

∴ Temperature difference of the body (K) = Internal temperature(K) - surface temperature(K) = (X + 279) - (X + 280)

   = X - X + 280 - 273 = 7 K.

∴Temperature difference of the body (K) = 7 K

Also for Fahrenheit, Convert the individual temperatures (Surface temperature and internal temperature) to Fahrenheit before calculating the temperature difference of the body.

We use , F = 1.8C + 32

Where C = temperature in Celsius.

also,

Let The Surface temperature Be = X °C

And the internal Temperature of the body will be = (X + 7) °C

Converting to Fahrenheit

Surface Temperature of the body = 1.8X + 32 °F

Internal Temperature of the body = 1.8(X+7) + 32 = 1.8X + 12.6 + 32

Internal Temperature of the body = 1.8X + 44.6 °F

∴ The temperature difference of the body (°F) = Internal temperature(°F) - surface temperature(°F) = (1.8X + 44.6) - (1.8X + 32)

      surface temperature(°F) = 1.8X - 1.8X  + 44.6 - 32

       surface temperature(°F) = 12. 6 °F.

   

3 0
2 years ago
A piano string sounds a middle A by vibrating primarily at 220 Hz.a)Calculate its period.b)Calculate its angular frequency.c)Cal
chubhunter [2.5K]

a) 4.5 ms

The period of a wave is given by:

T=\frac{1}{f}

where f is the frequency.

For the note in this problem, f = 220 Hz, so the period of the wave is

T=\frac{1}{f}=\frac{1}{220 Hz}=4.5\cdot 10^{-3} s = 4.5 ms

b) 1381.6 rad/s

The angular frequency is given by:

\omega=2 \pi f

where f is the frequency.

In this problem, f = 220 Hz, so the angular frequency is

\omega=2 \pi (220 Hz)=1381.6 rad/s

c) 1.1 ms

The frequency of the "high A" is four times the frequency of the piano string, so

f=4 \cdot 220 Hz=880 Hz

And so, its period is

T=\frac{1}{f}=\frac{1}{880 Hz}=1.1\cdot 10^{-3} s=1.1 ms

d) 5526.4 rad/s

The angular frequency is given by:

\omega=2 \pi f

where f is the frequency.

For this note, f = 880 Hz, so the angular frequency is

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5 0
1 year ago
The wind blows a jay bird south with a force of 300 Newtons. The
adoni [48]

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The  jay bird flies north, against the wind, with a force of 100  newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

F=\sqrt{300^2+100^2} \\\\F=316.22\ N

Hence, the net force on the jay bird is 316.22 N.

6 0
2 years ago
An 8.00 kg point mass and a 15.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be
Verdich [7]

Answer:a=2.23\times 10^{-9} m/s^2

Explanation:

Given

Mass of first Point mass m_1=8 kg

Mass of second Point m_2=15 kg

distance between them d=50 cm

third point mass m_3=m

Distance between m\ and\ m_1 is\ 20 cm

Distance between  m\ and\ m_2 is 30 cm

Force Due to m_1\ and\ m F_1=\frac{Gmm_1}{d_1^2}

F_1=\frac{8Gm}{(0.2)^2}

F_1=200 mG

F_2=m\frac{Gmm_2}{d_2^2}

F_1=m\frac{15Gm}{(0.3)^2}

F_2=166.67 mG

Net Force

F_{net}=F_1-F_2

=200 mG-166.67 mG

=33.33 mG

F_{net}=222.33\times 10^{-11} N

F_{net}=2.23m\times 10^{-9} N

acceleration a=\frac{2.23m\times 10^{-9}}{m}

a=2.23\times 10^{-9} m/s^2

towards 8 kg mass

3 0
1 year ago
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