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Luda [366]
3 years ago
11

Two points, point A and point B, are situated above a current-carrying wire. Point B is located at a distance R from the wire, w

hich is twice as far from the wire as point A. By what factor is the magnetic field at point A larger or smaller than the magnetic field at point B?
Physics
1 answer:
padilas [110]3 years ago
7 0

Explanation:

It is given that, there are two points situated above a current-carrying wire. Point B is located at a distance R from the wire, which is twice as far from the wire as point A such that,

r_B=R

And r_A=R/2

We know that the magnetic field at a distance R is inversely proportional to the distance from wire as :

B=\dfrac{\mu_o}{2\pi}\dfrac{I}{r}

B_A\propto\dfrac{1}{(R/2)}

B_B\propto\dfrac{1}{(R)}

\dfrac{B_A}{B_B}=\dfrac{1/(R/2)}{1/R}

\dfrac{B_A}{B_B}=\dfrac{2}{1}

B_A=2B_B

So, the magnetic field at point A larger than the magnetic field at point B by a factor of 2. Hence, this is the required solution.

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[b] it id heated from 4o

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What are the conditions of the thermonuclear fusion reaction in the sun?​
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2 years ago
An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
A person walks first at a constant speed of 5.50 m/s along a straight line from point A to point B and then back along the line
Sav [38]

Answer:

4.25 m/s

Explanation:

They walked the first distance at 5.50 m/s, then the same distance at 3 m/s.

Since the distances are equal, the average speed is simply the average of 5.50 and 3.

(5.50 + 3) / 2 = 4.25

Her average speed over the entire trip is 4.25 m/s.

8 0
3 years ago
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