Physics, 05.05.2020 08:15, myparentsrock17

A charged particle is accelerated from rest through a potential difference of magnitude |ΔV|. After exiting the potential difference at an emission point, the particle enters a region of uniform magnetic field. The magnetic field is perpendicular to the particle's velocity, and the particle travels along a complete circular path. The particle's mass is 2.10 ✕ 10−16 kg, its charge is 26.0 nC, and the magnetic field magnitude is 0.600 T. The particle's circular path, as it returns to the emission point, encloses a magnetic flux of 15.0 µWb.

a. What is the speed in (m/s) of the particle when it is in the region of the magnetic field?
b. What is the magnitude of the potential difference through which the particle was accelerated?

Answers: 2

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answered: lizzyhearts

The speed of the particle is v = 209485.71 m/s

The potential difference is $\Delta V = 177.2 \ V$

Explanation:

From the question we are told that

The mass of the particle is $m = 2.10 *10^{-16} kg$

The charge on the particle is $q = 26.0 nC = 26.0 *10^{-9} C$

The magnitude of the magnetic field is B = 0.600 T

The magnetic flux is $\O = 15.0 \mu Wb = 15.0 *10^{-6} Wb$

The magnetic flux is mathematically represented as

$\O = B *A$

Where A is the the area mathematically represented as

$A = \pi r^2$

Substituting this into the equation w have

$\O = B (\pi r^2 )$

Making r the subject of the formula

$r = \sqrt{\frac{\O}{B \pi} }$

Substituting value

$r = \sqrt{\frac{15 *10^{-6}}{3.142 * 0.6} }$

$r = 2.82 *10^{-3}m$

For the particle to form a circular path the magnetic force the partial experience inside the magnetic must be equal to the centripetal force of the particle and this is mathematically represented as

F_q = F_c

Where F_q = q B v

and $F_c = \frac{mv^2 }{r}$

Substituting this into the equation above

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

making v the subject

$v = \frac{r q B}{m}$

substituting values

$v = \frac{2.82 * 10^{-3} * 26 *10^{-9} 0.6}{2.10*10^{-16}}$

v = 209485.71 m/s

The potential energy of the particle before entering the magnetic field is equal to the kinetic energy in the magnetic field

This is mathematically represented as

PE = KE

Where $PE = q * \Delta V$

and $KE = \frac{1}{2} mv^2$

Substituting into the equation above

$q \Delta V = \frac{1}{2} mv^2$

Making the potential difference the subject

$\Delta V = \frac{mv^2}{2 q}$

$\Delta V = \frac{2.16 * 10^ {-16} * (209485.71)^2 }{2 * 26 *10^{-9}}$

$\Delta V = 177.2 \ V$

answered: Guest

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

answered: Guest

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A charged particle is accelerated from rest through a potential difference of magnitude |ΔV|. After...