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lianna [129]
3 years ago
10

A particle that has mass m and charge q enters a uniform magnetic field that has magnitude B and is directed along the x axis. T

he initial velocity of the particle is in the xy plane.
a) Describe the particle's path as it travels through the magnetic field.

b) If the particle enters the magnetic field at t = 0, what is its angular displacement at t=2{eq}\pi {/eq}m/qB? Express your answer in terms of some or all of the variables m, q, B, and the constant {eq}\pi {/eq}?
Physics
1 answer:
Usimov [2.4K]3 years ago
6 0

Answer:

a) The trajectory will be a helical path.

b) θ = 2*π rad

Explanation:

a) Since the initial velocity of the particle has a component parallel (x-component) to the magnetic  field B , the trajectory will be a helical path.

b) Given

t = 2*π*m/(q*B)

We can use the equation

θ = ω*Δt

where

θ is the angular displacement

ω is the angular speed, which is obtained as follows:

ω = q*B/m

then we have

θ = (q*B/m)*2*π*m/(q*B)

⇒  θ = 2*π rad

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The electric potential at the origin of the xy coordinate system is negative infinity

<h3>What is the electric field due to the 4.0 μC charge?</h3>

The electric field due to the 4.0 μC charge is E = kq/r² where

  • k = electric constant = 9.0 × 10 Nm²/C²,
  • q = 4.0 μC = 4.0 × 10 C and
  • r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m

<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

  • k = electric constant = 9.0 × 10 Nm²/C²,
  • q' = -4.0 μC = -4.0 × 10 C and
  • r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m

Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is

E" = E + E'

= -2E

= -2kq/r²

<h3>What is the electric potential at the origin?</h3>

So, the electric potential at the origin is V = -∫₂⁰E".dr

= -∫₂⁰-2kq/r².dr

Since E and dr = dx are parallel and r = x, we have

= -∫₂⁰-2kqdxcos0/x²

= 2kq∫₂⁰dx/x²

= 2kq[-1/x]₂⁰

= -2kq[1/x]₂⁰

= -2kq[1/0 - 1/2]

= -2kq[∞ - 1/2]

= -2kq[∞]

= -∞

So, the electric potential at the origin of the xy coordinate system is negative infinity

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You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius
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Answer:

t = 0.0735 m

Explanation:

Angular acceleration of the flywheel is given as

\alpha = 3 rad/s^2

now after t = 8 s the speed of the flywheel is given as

\omega = \alpha t

\omega = 3 \times 8

\omega = 24 rad/s

now rotational kinetic energy of the wheel is given as

K = \frac{1}{2}I\omega^2

K = \frac{1}{2}(\frac{1}{2}mR^2)(24^2)

800 = \frac{1}{4}m(0.23)^2(24^2)

m = 105 kg

now we have

m = \rho (\pi R^2) t

105 = 8600(\pi \times 0.23^2) t

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Introduced species often thrive and multiply in an environment very different from their original one. Why are they often able t
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The average adult in the US watches _____ hours of television each week.
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A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

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

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

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