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professor190 [17]
3 years ago
7

a proton is moving horizontally at 7.9 x 10^5 m/s. It passes through a vertical magnetic field that points downward. In which di

rection is the proton deflected by the magnetic field? A. up B. to the right of its velocity C. down D. to the left of its velocity

Physics
1 answer:
blondinia [14]3 years ago
7 0
There are four key characteristics of the magnetic force on a moving charge.

First, Its magnitude is proportional to the magnitude of the charge. 
Second, the magnitude of the force is also proportional
Third, the magnetic force depends on the particle’s
velocity. 
Fourth, we find by experiment that the magnetic force \vec{F} does not have the same direction as the magnetic field \vec{F} but instead is always perpendicular to both \vec{B} and the velocity \vec{v} 

So, in a mathematical language this is given by:

\vec{F}=q\vec{v} \times \vec{B}

So, this is a cross product. Therefore, applying the Right-hand rule:

The answer is:

<span>D. to the left of its velocity.

Explaining this in other words:

"The force </span>\vec{F} is directed into the plane of the paper, so this is the direction of the proton"

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leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

5 0
3 years ago
Suppose a magnetic reversal accoured today .
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3 years ago
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Answer:

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Answer:

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The answer is D using the work formula
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