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2 years ago

8

An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector

inside a TV tube. The electron strikes the screen after traveling 11 cm horizontally and 34 cm vertically upward with no horizontal acceleration. What is the constant vertical acceleration provided by the deflector? (The effects of gravity can be ignored.)

1 answer:

Ghella [55]2 years ago

7 0

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

$v_{x}$ = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

$t = \frac{X}{v_{x}}$

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel = 3.67 x 10⁻⁸ sec

$v_{y}$ = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = $v_{y}$ t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

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Answer:

The other two small angles are 45° each

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A long, straight wire carrying a current of 3.45 A moves with a constant speed v to the right. A 5-turn circular coil of diamete

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Answer:

I = 69.3 μA

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Current through the straight wire, I = 3.45 A

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Resistance of the coil, $R = 3.25 \mu ohms$

Distance of the wire from the center of the coil, d = 20 cm = 0.2 m

The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.

$B_{1} = \frac{\mu_{0}I }{2\pi d}$

$B_{1} = \frac{4\pi* 10^{-7} *3.45 }{2\pi *0.2}\\B_{1} =0.00000345 T$

Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil

$B_{2} = \frac{\mu_{0}I }{2\pi(2d)) } \\B_{2} = \frac{\mu_{0}I }{4\pi d } \\$

$B_{2} = \frac{4\pi* 10^{-7} *3.45 }{2\pi *2*0.2}\\B_{2} = 0.000001725 T$

Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345

ΔB = -0.000001725

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E = -N (Δ∅)/Δt

Δ∅ = A ΔB

Area, A = πr²

diameter, d = 0.0125 m

Radius, r = 0.00625 m

A = π* 0.00625²

A = 0.0001227 m²

Δ∅ = -0.000001725 * 0.0001227

Δ∅ = -211.6575 * 10⁻¹²

E = -N (Δ∅)/Δt

$E = -5\frac{-211.6575 * 10^{-12} }{4.70} \\E = 225.17 * 10^{-12} V$

Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms

I = E/R

$I = \frac{225.17 * 10^{-12} }{3.25 * 10^{-6} }$

I = 0.0000693 A

I = 69 .3 * 10⁻⁶A

I = 69.3 μA

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