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If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f

dsp73

1. The magnitude of the magnetic field doubles

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

where $\mu_0$ is the vacuum permeability, I is the current in the wire, r is the distance from the wire.

As we see from the formula, the intensity of the magnetic field is directly proportional to the current: if the current increases from 5 A to 10 A, it means it doubles, so the magnetic field doubles as well.

2. The magnitude of the magnetic field halves

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

We see that the magnitude of the magnetic field is inversely proportional to the distance from the wire (r). In this case, the distance of the particle is changed from 10 cm to 20 cm, so it is doubled: therefore, the magnitude of the field will become half of the initial value.

3. The force reverses direction

Explanation: the force exerted on a charged particle in a magnetic field is:

$F=qvB sin \theta$

where q is the charge, v is the speed of the particle, B is the magnetic field intensity and $\theta$ the angle between the direction of v and B. If the charge of the particle is switched from 2 µC to –2µC, the magnitude of the force does not change (because the absolute value of q does not change), however the charge q gets a negative sign (-), so the sign of the force changes and gets a negative sign too, so the force reverses direction.

7 0

3 years ago

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A car is moving 18 m/s to the east. if it takes the car 5 seconds to reach a velocity of 19 m/s to the east, what is its acceler

almond37 [142]

<span>C. 0.2 m/s2 east

Hope it helps!
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5 0

3 years ago

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A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m

Irina18 [472]

The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.

7 0

3 years ago

Three resistors of 100 W, 3900 W, and 1000 W are connected in series across a 200-V battery. What is the voltage drop across the

Gala2k [10]

Answer:

40 V

Explanation:

I will assume that the resistors are

100 and 3900 and 1000 OHMS <=====(NOT W)

In series , the resistances add together 100 + 3900 + 1000 = 5000 ohms total

V = IR

I = V / R so the total current will be 200 v / 5000 ohms = .04 amps

this is the current through all of the resistors

so for the 1000 ohm resistor V = IR .04 (1000) = 40 V

7 0

2 years ago

A 2,000 g quantity of C-14 is left to undergo radioactive decay. The half-life of C-14 is approximately 5,700 years. After going

Katyanochek1 [597]

Start with 2,000 grams.
After 1 half-life, 1,000 grams are left.
After another half-life, 500 are left.
After another half-life, 250 are left.
After another half-life, 125 are left.

That was FOUR half-lifes.

X = 4 .

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3 years ago

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