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Fantom [35]
3 years ago
5

You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current i

n the wire is 1.2A. What is the radius of the coils and how many loops, N, are there
Physics
1 answer:
Anni [7]3 years ago
8 0

Answer:

<em>radius of the loop =  7.9 mm</em>

<em>number of turns N ≅ 399 turns</em>

Explanation:

length of wire L= 2 m

field strength B = 3 mT = 0.003 T

current I = 12 A

recall that field strength B = μnI

where n is the turn per unit length

vacuum permeability μ  = 4\pi *10^{-7}  T-m/A = 1.256 x 10^-6 T-m/A

imputing values, we have

0.003 = 1.256 x 10^−6 x n x 12

0.003 = 1.507 x 10^-5 x n

n = 199.07 turns per unit length

for a length of 2 m,

number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>

since  there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

this length is also equal to the circumference of each loop

the circumference of each loop = 2\pi r

0.005 = 2 x 3.142 x r

r = 0.005/6.284 = 7.9*10^{-4} m = 0.0079 m =<em> 7.9 mm</em>

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Answer:Zero

Explanation:

Given

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8 0
3 years ago
If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i
kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from 15 cm to 21 cm will be , E_{1}=27J

The energy used for stretch the spring from 21 cm to 27 cm will be , E_{2}=45J

using the energy of spring formula ,we find that

27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))

45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))

Dividing both the equation will get,

\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm

Therefore, the natural length of the spring is, 9 cm.

4 0
3 years ago
The atmosphere of Mercury and Mars are very thin. What effect does the thin atmosphere have on the temperature on the surface of
KengaRu [80]

Answer:

Very hot during the day and very cold at night.

Explanation:

Due to the thin atmosphere, they have very hot climate during the day time and very cold climate at night. This happens because they contain very low amounts of greenhouse gases. These gases retain the heat at night. The atmosphere also prevents excessive light and UV rays from entering. The thin  atmosphere leads to many asteroids and comets hitting the surface of the planet. On earth, these asteroids usually, burn up in the mesosphere layer of the atmosphere. These asteroid collisions cause massive fires. This in turn,  causes the temperature to increase during the day. During the night time, massive fires cannot burn due to the low temperature because of the lack of greenhouse gases.

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3 years ago
A student is watching waves come in from the ocean. He noticed that the first wave he saw (Wave A) had twice the amplitude of th
Alika [10]

Answer:

Wave A

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8 0
3 years ago
Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?
Stolb23 [73]

Answer:

200 cal/^{\circ}C

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

Q=C\Delta T

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Q is the heat supplied

C is the heat capacity of the object

\Delta T is the change in temperature

In this problem we have:

Q=10,000 cal is the energy supplied

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Therefore, the heat capacity of the object is:

C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C

6 0
3 years ago
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