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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 square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )

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(c)

R = 3 ohm

e = -\frac{d\phi}{dt}

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e = 1.89 x 10^-7 V

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