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iren2701 [21]
3 years ago
11

A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil h

as 100 turns how much current must pass through the coil?
Physics
1 answer:
tresset_1 [31]3 years ago
6 0

Answer:

The current pass through the coil is 6.25 A

Explanation:

Given that,

Diameter = 25 cm

Magnetic field = 1.0 mT

Number of turns = 100

We need to calculate the current

Using the formula of magnetic field

B =\dfrac{\mu_{0}NI}{2\pi r}

I=\dfrac{B\times2\pi r}{\mu N}

Where, N = number of turns

r = radius

I = current

Put the value into the formula

I=\dfrac{1.0\times10^{-3}\times2\pi\times12.5\times10^{-2}}{4\pi\times10^{-7}100}

I=6.25\ A

Hence, The current passes through the coil is 6.25 A

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A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located
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The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁  is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241  \times 10^{-19 } \ C

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

#SPJ2

6 0
2 years ago
What type of fire can be put out safely with water?
nignag [31]
Flame of fire could get put out with water
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3 years ago
Devise an exponential decay function that fits the given​ data, then answer the accompanying questions. Be sure to identify the
7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

P=P_0e^{-kh}

where,

P_0 =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

P=\dfrac{1}{3}P_0

\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}

Now

P=\dfrac{1}{2}P_0

ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft

The altitude will be 22145.27733 ft

P=0.02P_0

0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft

The elevation is 124984.76055 ft

6 0
3 years ago
An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi
elena55 [62]

Answer:

a) \vec F_{B} = 8.766\times 10^{-14}\,T\,k, b) \vec F_{B} = -8.766\times 10^{-14}\,T\,k

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

\vec F_{B} = q\cdot \vec v \times \vec B

The charge of the electron is equal to -1.602\times 10^{-19}\,C. Then, cross product can be solved by using determinants:

\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}

The magnetic force is:

\vec F_{B} = 8.766\times 10^{-14}\,T\,k

b) The charge of the proton is equal to 1.602\times 10^{-19}\,C. Then, cross product has the following determinant:

\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}

The magnetic force is:

\vec F_{B} = -8.766\times 10^{-14}\,T\,k

8 0
2 years ago
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