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pogonyaev
3 years ago
10

How would I explain part a?

Physics
1 answer:
grandymaker [24]3 years ago
3 0

Electric field on the axis of ring on both sides is along the axial line

This electric field is away from the center on both sides

Now when an electron is placed on the axis of the ring then we will see that it will experience the electrostatic force on the electron.

The direction of force is always towards the center of the ring because here electron is a negatively charged particle and it will move opposite to the direction of electric field.

So as we release the electron from a distance of 30 cm the electron will move towards the center of ring and accelerate to its maximum speed till it will reach to the center of the ring

After that the electron will reach to the center of the ring and then moves out on the opposite side and then decelerate to zero speed as it will reach on other side.

Now after this it will again move towards the center of the ring and continue its motion similar way

So this is an oscillating motion about the center of the ring.

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Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R =  60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀  = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

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T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

I(t) = I_o(1-e^{-\frac{t}{T}})

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

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