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the picture tube in a television uses magnetic deflection coils rather than electric deflection plates. suppose an electron beam is accelerated through a 55.0 -kv potential difference and then through a region of uniform magnetic field 1.00 cm wide. the screen is located 6.0 cm from the center of the coils and is 60.0 cm wide. when the field is turned off, the electron beam hits the center of the screen. what field magnitude is necessary to deflect the beam to the side of the screen? ignore relativistic corrections.



Answer :

The field magnitude required to divert the beam away from the screen is 3.1588 × [tex]10^{-4}[/tex].

A magnetic force is applied to a charged particle when it enters a magnetic field. This force's strength is determined by the particle's speed, magnetic field, and charge density. Fleming's Left Hand Rule determines the force's direction. The component of the velocity transverse to the magnetic field provides the required centripetal force if the charged object arrives at an angle.

V = 55.0 kV

d = 1.00cm

Tan∅ = ((1 ÷ 2) × 60) ÷ 6.0 = 5

∅ = 26.6°

Sin∅ = 1 ÷ R

R = 1 ÷ Sin(26.6°)

R = 1 ÷ 0.44

R = 2.27

Total work completed equals Kinectic Energy change

1 ÷ 2 mv² = qΔV

0.5 × 9.1 × [tex]10^{-31}[/tex] × v² = 1.6 × [tex]10^{-19}[/tex] × 55.0 × [tex]10^{3}[/tex]

v² = (8.8 × [tex]10^{-15}[/tex]) ÷ (4.55 × [tex]10^{-31}[/tex])

v² = 1.93 × [tex]10^{16}[/tex]

v = √1.93 × [tex]10^{16}[/tex]

v = 138924439

v = 1.3 × [tex]10^8[/tex]

The magnetic force produces the centripetal force.

qvB = mv² ÷ R

B = (mv ÷ qR)

B = (9.1 × [tex]10^{-31}[/tex] × 1.3 × [tex]10^8[/tex]) ÷ (1.6 × [tex]10^{-19}[/tex] × 2.27)

B = (11.83 × [tex]10^{-23}[/tex]) ÷ (3.745 × [tex]10^{-19}[/tex])

B = 3.1588 × [tex]10^{-4}[/tex]

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