Question

43. In an electron gun of a TV set an electron with an initial speed of 1000 m/s enters a region where it is electricallyaccelerated it emerges out of this region after 1 micro second with a speed of 4*10^5 m/s Maximum Length (in meters)of electron gun and acceleration (in m/s^2) of the electron are? ​

Answer 1

Answer:

Kinetic energy of electron, K=

2

1

mv

2

=2keV

Speed of electron, v=

m

2K

=

9.1×10

−31

2×2×1.6×10

−16

ms

−1

=2.65×10

7

ms

−1

Since, the velocity (

v

) of the electron makes an angle of θ=60

o

with the magnetic field

B

, the path will be a helix.

So, the particle will hit S if Gs=np; Here, n=1,2,3,....

P= pitch of helix =

qB

2πm

vcosθ

⇒GS=

qB

n2πmvcosθ

⇒B=

q(GS)

n2πmvcosθ

But for B to be minimum, n=1

B

min

=

q(GS)

2πmvcosθ

Substituting the values, we have:

B

min

=

(1.6×10

−19

)(0.1)

(2π)(9.1×10

−31

)(2.67×10

7

)(

2

1

)

B

min

=4.73×10

−3

T

Answer 2

Answer:

0.2 metres, 399×1000000000m/s2