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alexandr1967 [171]
3 years ago
12

How can the motion of a charged particle be used to distinguish between a magnetic field and an electric field in a certain regi

on? The charged particle follows a spiral path in a magnetic field. The charged particle follows a linear path in a magnetic field. The charged particle follows a spiral path in an electric field
Physics
1 answer:
Helen [10]3 years ago
4 0

Answer:

The charged particle follows a spiral path in a magnetic field.

Explanation:

A charge immersed in a region with an electric field experiences a force that acts along the same direction of the electric field. In particular:

- The force has the same  direction as the electric field if the charge is positive

- The force has the opposite direction as the electric field if the charge is negative

Therefore, a charge moving in an electric field is accelerated along the direction of the electric field.

On the other hand, a charge in motion in a region with a magnetic field experiences a force that acts perpendicular to the direction of the field. This means that a charge in motion in a magnetic field will acquire a circular motion in the plane perpendicular to the direction of the magnetic field.

As a result, if the particle has also a original motion outside this plane, its final motion will consist of:

- A uniform motion along that direction, +

- A circular motion along the plane perpendicular to the field

So, the resultant motion of the particle will be a spiral path. So the correct answer is

The charged particle follows a spiral path in a magnetic field.

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A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
3 years ago
A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

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I believe it is call “Acceleration”
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