Physics, 03.03.2020 23:19 miya214

A deuteron, with the same charge but twice the mass of a proton, moves with a speed of 6.00 × 105 m/s perpendicular to a uniform magnetic field of 0.0525 T. Which of the paths described below would it follow? (qp = 1.60 × 10−19 C and md = 3.34 × 10−27 kg)

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Answer from: mistermansour07

Explanation:

The given data is as follows.

v = $6.00 \times 10^{5} m/s$

B = 0.0525 T, q = $1.60 \times 10^{-19}$

m = $3.34 \times 10^{-27} kg$

It is known that relation between mass and magnetic field is as follows.

$\frac{mv^{2}}{r} = Bvq$

or, r = $\frac{mv^{2}}{Bvq}$

So, putting the given values into the above formula and we will calculate the radius as follows.

r = $\frac{mv^{2}}{Bvq}$

= $\frac{3.34 \times 10^{-27} kg \times 6.00 \times 10^{5} m/s}{0.0525 T \times 1.60 \times 10^{-19}}$

= $\frac{20.04 \times 10^{-22}}{0.084 \times 10^{-19}}$

= 0.238 m

Thus, we can conclude that radius of the circular path is 0.238 m.

Answer from: cam1976

The radius is $238.57\times10^{-3}\ m$

Explanation:

Given that,

Speed $v=6.00\times10^{5}\ m/s$

Magnetic field = 0.0525 T

We need to calculate the radius

Using relation of centripetal force and magnetic force

F=qvB

$\dfrac{mv^2}{r}=qvB$

$r=\dfrac{mv^2}{qvB}$

$r=\dfrac{mv}{qB}$

Put the value into the formula

$r=\dfrac{3.34\times10^{-27}\times6.00\times10^{5}}{1.60\times10^{-19}\times0.0525}$

$r=0.238\ m$

$r=238.57\times10^{-3}\ m$

Hence, The radius is $238.57\times10^{-3}\ m$

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

The answer should be a.

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A deuteron, with the same charge but twice the mass of a proton, moves with a speed of 6.00 × 105 m/...

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