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n200080 [17]
3 years ago
9

An electron travels with speed 6.0×10^6 m/s between the two parallel charged plates shown in the figure. The plates are separate

d by 1.0 cm and are charged by a 200 Volt battery.
What magnetic field strength will allow the electron to pass between the plates without being deflected?
Physics
1 answer:
svlad2 [7]3 years ago
5 0

Answer:

B= 3.33 m T

Explanation:

Given that

Speed ,C= 6 x 10⁶  m/s

d= 1 cm = 0.01 m

V= 200 V

The electric field E given as

V= E .d

E=Electric field

d=Distance

V=Voltage

200 = 0.01 x E

E=20000 V/m

The relationship between magnetic and electric field given as

E= C x B

20000 = 6 x 10⁶   x B

B =3333.333 x 10⁻⁶ T

B= 3.33 x 10⁻³ T

B= 3.33 m T

Therefore the magnetic filed will be 3.33 m T.

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A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
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Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

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Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

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Answer:

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Explanation:

Kinetic energy is defined as:

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