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3 years ago

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A proton is accelerated through a potential difference of 250V. It then enters a uniform magnetic field and moves in a circular

path of radius 10.0cm. What is the magnitude of the magnetic field? What is the period of revolution for the circling proton?

2 answers:

3241004551 [841]3 years ago

6 0

Answer: a) 7.52 ×10^-4 T, b) 4.72×10^-7s

Explanation: by using the work energy thereom, the work done on the electron by the voltage source (potential) equals the kinetic energy.

qV =mv^2/2

Where q = magnitude of electronic charge = 1.609×10^-19 c

V = potential difference = 250v

m = mass of an electronic charge = 9.11×10^-31 kg

v = velocity of electron =?

By substituting the parameters, we have that

1.609×10^-19 × 250 = 9.11×10^-31 × v^2/2

1.609×10^-19 × 250 ×2 = 9.11×10^-31 ×v^2

v^2 = 1.609×10^-19 × 250 ×2/ 9.11×10^-31

v^2 = 8.05×10^-17/ 9.11×10^-31

v^2 = 1.77×10^14

v = √1.77×10^14

v = 1.33×10^7 m/s

The centripetal force required for the motion of the electron is gotten from the magnetic force on the electron.

qvB = mv^2/r

By cancelling "v" on both sides of the equation, we have that

qB = mv/r

Where r = radius = 10cm = 0.1m

1.609×10^-19 × B = 9.11 ×10^-31 ×1.33×10^7/ 0.1

B = (9.11 ×10^-31 × ×1.33×10^7)/ (0.1 ×1.609×10^-19)

B = 1.21×10^-23/ 1.609×10^-20

B = 0.752×10^-3

B = 7.52 ×10^-4 T

To get the period of motion, we recall that

v = ωr

Where ω = angular frequency

1.33×10^7 = ω×0.1

ω = 1.33×10^7/ 0.1

ω = 13.3×10^7

ω = 1.33×10^6 rad/s

But the period (T) of a periodic motion is defined as

T = 2π/ω

T = 2 × 3.142 / 1.33×10^6

T = 4.72×10^-7s

Lyrx [107]3 years ago

6 0

Answer:

B = 0.0228T

T= 2.87×10-⁶s

Explanation:

Given V = 250V, R = 10cm = 0.1m

q = 1.6×10-¹⁹C, m = 1.67×10-²⁷kg

From the work energy theorem the electric energy of the field = kinetic energy of the proton.

qV = 1/2×mv²

v = √(2qV/m) = √(2×1.6×10-¹⁹×250/(1.67×10-²⁷))

v = 218870m/s

The magnetic force on the proton by newton's 2nd law of motion is related as

qvB = ma = m× v²/R

Rearranging

B = mv/qR = 1.67×10-²⁷ × 218870/(1.6×10-¹⁹ × 0.1) = 0.0228T

T = 2π/ω = 2πR/v = 2π×0.1/218870 = 2.87×10-⁶ = 2.87μs.

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A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

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Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

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Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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