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Stolb23 [73]
3 years ago
12

A single-turn circular loop of radius 14 cm is to produce a field at its center that will just cancel the earth's magnetic field

at the equator, which is 0.7 g directed north. the permeability of free space is 4 π × 10−7 t · m/a. find the current in the loop. answer in units of a
Physics
1 answer:
Olin [163]3 years ago
6 0

Answer:

49 A

Explanation:

We are given that

Number of turns,N=1

Radius,r=14 cm=\frac{14}{100}=0.14 m

1 m=100 cm

Magnetic field,B=0.7 G=0.7\times 10^{-4} T

1G=10^{-4} T

\mu_0=4\pi\times 10^{-7} Tm/A

We know that magnetic field

B=\frac{\mu_0IN}{2\pi r}

Substitute the values

0.7\times 10^{-4}=\frac{4\pi\times 10^{-7}\times 1\times I}{2\pi\times 0.14}

I=\frac{0.7\times 10^{-4}\times 2\pi\times 0.14}{4\pi\times 10^{-7}\times 1}

I=49A

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1 year ago
What could be used as another word for electrical potential
tekilochka [14]

Answer:

hey mate

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Explanation:

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5 0
2 years ago
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Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th
Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

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Answer:

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