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Answer 1

$\sf\huge\bold{Answer:}$

Given :

Magnetic field = 0.05 G

Monoenergetic electron beam = 18 keV

Distance = 30cm

$m_{e} = 9.11 \times {10}^{ - 31} C$

To find :

Deflection of beam

Solution :

Velocity of the beam can be found from its kinetic energy.

$= \frac{1}{2} mv {}^{2}$

$\implies \: v = \sqrt{ \frac{2E}{m} }$

$= \sqrt{ \frac{2 \times 18 \times {10}^{3} \times 1.6 \times {10}^{ - 19} }{9.1 \times {10}^{ - 31} } }$

$= 0.795 \times {10}^{ 8} ms {}^{ - 1}$

The magnetic force acting on the beam tends it to follow a circular path. The magnetic force equals the centripetal force.

$\implies \: qvB = \frac{mv {}^{2} }{r}$

$\implies \: r = 11.3m$

The up and down deflection of the electron beam is

$x = r(1 - \cos \: \theta)$

$\sin( \theta) = \frac{d}{r} = \frac{0.3}{11.3}$

$\implies \: \theta \: = 1.521 \degree$

$\therefore \:$

deflection

$x = 11.3(1 - \cos(1.521 \degree)$

$\implies \: x = 3.9mm$

Answer 2

Answer:

Velocity of the beam can be found from its kinetic energy.

=

2

1

mv

2

⟹v=

m

2E

=

9.1×10

−31

2×18×10

3

×1.6×10

−19

=0.795×10

8

m/s

The magnetic force acting on the beam tends it to follow a circular path. The magnetic force equals the centripetal force.

qvB=

r

mv

2

⟹r=11.3m

The up and down deflection of the electron beam is

x=r(1−cosθ)

sinθ=

r

d

=

11.3

0.3

⟹θ=1.521

∘

Therefore deflection x=11.3(1−cos1.521

∘

)=3.9mm

Explanation:

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