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vitfil [10]
3 years ago
6

An electron moves through a region of crossed electric and magnetic fields. The electric field E = 3059 V/m and is directed stra

ight down. The magnetic field B = 1.14 T and is directed to the left. For what velocity v of the electron into the paper will the electric force exactly cancel the magnetic force?
Physics
1 answer:
dangina [55]3 years ago
7 0

Answer:

v = 2683.33 m/s

Explanation:

The magnetic force and the electric force on the electron must be the same, in order for them to cancel each other:

Electric\ Force = Magnetic\ Force\\Eq = qvBSin\theta \\\\v = \frac{E}{BSin\theta}

where,

v = velcoity of electron = ?

E = Electric Field = 3059 V/m

B = Magnetic Field = 1.14 T

θ = Angle between velocity and magnetic field = 90°

Therefore,

v = \frac{3059\ V/m}{(1.14\ T)Sin90^o}

<u>v = 2683.33 m/s</u>

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                   v(t) = \frac{dy}{dt}

                v(t) = A \omega cos(kx + \omega t)

        v(t) = (0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)

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The acceleration of the particle in the location is

            a(t) = \frac{dv}{dt}

           a(t) = -A \omega 2sin(kx + \omega t)

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           a(t) = -9.49 m/s^{2}

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 m/s^{2} .

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               \lambda = \frac{2\pi}{k}

              \lambda = (frac{2\pi}{\frac{\pi}{11})&#10;

              \lambda = 22 m

The period of the wave is

             T = \frac{2 \pi}{\omega}

             T = \frac{2 \pi}{4 \pi}

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Now, we will calculate the speed of propagation of wave as follows.

                    v = \frac{\lambda}{T}

                       = \frac{22 m}{0.5 s}

                       = 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

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