Physics, 06.05.2020 06:36 homework1911

A time-dependent but otherwise uniform magnetic field of magnitude Bo(t) is confined in a cylindrical region of radius 6.5 cm. Initially the magnetic field in the region is pointed out of the page and has a magnitude of 3.5 T, but it is decreasing at a rate of 29.5 G/s. Due to the changing magnetic field, an electric field will be induced in this space which causes the acceleration of charges in the region.

Required:
What is the direction of acceleration of a proton placed in at the point 1.5 cm from the center?

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Answer from: rosebelll4555

The acceleration is a = 9197.11 m/s^2

Explanation:

From the question we are told that

The radius of the cylinder is $r = 6.5 \ cm = \frac{6.5}{100} = 0.065 m$

The magnitude of the magnetic field is B_i = 3.5T

The rate of decrease is $\frac{dB}{dt} = 29.5 \ G/s = \frac{29.5}{10000} = 29.5 * 10 ^{-4} T/s$

The distance from the center is $D = 1.5\ cm = \frac{1.5}{10}=0.15m$

Faraday's law of induction states mathematically that

$\epsilon = \frac{d \o}{dt}$

$\epsilon = \int\limits {E} \, dl$

Where $\epsilon$ is the induced emf

E is the magnitude of the electric field

i the change in length

$d \o$ is the change in magnetic flux which is mathematically represented as

$\o = BA$

substituting this into the above equation

$\int\limits {E} \, dl = \frac{d(BA)}{dt }$

$E l = A \frac{dB}{dt}$

Where l is the circumference of the circular loop formed in the cylinder which is mathematically represented as

$l = 2\pi r$

And A is the area of the circular loop formed which is mathematically represented as

$A = \pi r^2$

$E (2 \pi r ) = (\pi r^2 ) \frac{dB}{dt}$

$E = \frac{r}{2} \frac{dB}{dt}$

Substituting value

$E = \frac{0.065}{2} * 29.5 *10^{-4}$

$= 9.588*10^{-5} \ V/m$

Generally acceleration is mathematically represented as

$a = \frac{F}{m}$

Now F is the electric force which is mathematically represented as

F = qE

Substituting this into the question

$a = \frac{qE }{m }$

Where q is the charge on the proton with a constant value of

$q = 1.60 *10^ {-19 } C$

and m is the mass of the proton with a constant value of

$m = 1.67 *10^{-31} kg$

Substituting values

$a = \frac{1.60*10^{-19} * 9.588 *10^{-5}}{1.67 *10^{-27}}$

a = 9197.11 m/s^2

Answer from: Quest

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Answer from: Quest

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A time-dependent but otherwise uniform magnetic field of magnitude Bo(t) is confined in a cylindrica...

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