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4 years ago

7

You are designing a new solenoid and experimenting with material for each turn. The particular turn you are working with is a ci

rcular loop of radius 3.00 cm that carries a current of 12.0 A . Calculate the magnetic field B at the center of the loop.

1 answer:

andriy [413]4 years ago

8 0

Answer:

The magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.

Explanation:

Given;

Radius of circular loop, R = 3.00 cm = 0.03 m

Current in the loop, I = 12.0 A

Magnetic field at the center of circular loop is given as;

B = μ₀I / 2R

Where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

R is the radius of the circular loop

I is the current in the loop

Substitute the given values in the above equation and calculate the magnitude of the magnetic field;

B = (4π x 10⁻⁷ x 12)/ 0.03

B = 5.0272 x 10⁻⁴ T

Therefore, the magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.

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Assume that both the supply of and the demand for a good are relatively price elastic. The imposition of a per-unit excise tax o

sergey [27]

Answer:P:Increase Q:Decrease

Explanation: The price of the product will increase as the imposed per-unit excise tax on the sale of the good. This is true for both Perfectly-elastic and relatively elastic products.

The quantity demanded of the product will decrease as the price of the product increases, this his is true for both Perfectly-elastic and relatively elastic products. It will follow the law of Demand and Supply.

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4 years ago

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Alexus [3.1K]

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3 years ago

Alcohol of mass 21g and density 0.7gcm-3 is mixed with 10g of water. Determine the density

IRINA_888 [86]

Answer:

0.775 g/cm³

Explanation:

The given parameters are;

The mass of the alcohol = 21 g

The density of the of the alcohol = 0.7 g/cm³

The mass of water with whichit is mixed = 10 g

The density of water = 1.0 g/cm³

Therefore, the mass of the mixture, m = 21 g + 10 g = 31 g

Volume = Mass/Density

The volume of the alcohol = 21 g/(0.7 g/cm³) = 30 cm³

The volume of the water = 10 g/(1.0 g/cm³) = 10 cm³

Therefore, the volume of the mixture, V = 30 cm³ + 10 cm³ = 40 cm³

The density of the mixture, ρ = (Mass, m)/(Volume, V) = m/V

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The density of the mixture, ρ = 0.775 g/cm³.

3 0

3 years ago

two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will h

Aneli [31]

Answer:

On that line segment between the two charges, at approximately $0.7\; \rm m$ away from the smaller charge (the one with a magnitude of $5 \times 10^{-19}\; \rm C$ ,) and approximately $1.3\; \rm m$ from the larger charge (the one with a magnitude of $20 \times 10^{-19}\; \rm C$ .)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let $k$ denote the Coulomb constant, and let $q$ denote the size of a point charge. At a distance of $r$ away from the charge, the electric field due to this point charge will be:

$\displaystyle E = \frac{k\, q}{r^2}$ .

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let $q_1 = 5\times 10^{-19}\; \rm C$ and $q_2 = 20 \times 10^{-19}\; \rm C$ . Assume that the electric field is zero at $r$ meters to the right of the $5\times 10^{-19}\; \rm C$ point charge. That would be $(2 - r)$ meters to the left of the $20 \times 10^{-19}\; \rm C$ point charge. (Since this point should be between the two point charges, $0 < r < 2$ .)

The electric field due to $q_1 = 5\times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}$ .

The electric field due to $q_2 = 20 \times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}$ .

Note that at all point in this section, the two electric fields $E_1$ and $E_2$ will be acting in opposite directions. At the point where the two electric fields balance each other precisely, $| E_1 | = | E_2 |$ . That's where the actual electric field is zero.

$| E_1 | = | E_2 |$ means that $\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}$ .

Simplify this expression and solve for $r$ :

$\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0$ .

$\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0$ .

Either $r = -2$ or $\displaystyle r = \frac{2}{3}\approx 0.67$ will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, $0 < r < 2$ . Therefore, $(-2)$ isn't a valid value for $r$ in this context.

As a result, the electric field is zero at the point approximately $0.67\; \rm m$ away the $5\times 10^{-19}\; \rm C$ charge, and approximately $2 - 0.67 \approx 1.3\; \rm m$ away from the $20 \times 10^{-19}\; \rm C$ charge.

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3 years ago

A 0.187 A current flows through a wire. How much time will it take for 2.00 C of charge to flow past a point in the wire?

Oksana_A [137]

Answer:A 0.187 A current flows through a wire. How much time will it take for 2.00 C of charge to flow past a point in the wire?

A. 10.7 s

B. 2.18 s

C. 0.374 s

D. 0.0935 s

Explanation:

A 0.187 A current flows through a wire. How much time will it take for 2.00 C of charge to flow past a point in the wire?

A. 10.7 s

B. 2.18 s

C. 0.374 s

D. 0.0935 s

7 0

3 years ago