Moving a magnet inside of a coil of wire will induce a voltage in the coil. How is the voltage in the coil increased?

My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 m from the ring's center. If

dsp73

Answer:

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Explanation:

Given,

Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
Radius of the ring = R = 0.70 m
Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring = $Q\ =\ \rho\times 2\pi R$

Potential energy due to the charged ring to the point on the x-axis is

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}\\$

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2\\$

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

$\therefore P.E.\ =\ K.E.\\\Rightarrow \dfrac{KQ}{\sqrt{R^2\ +\ x^2}}\ =\ \dfrac{1}{2}m_ev^2\\\Rightarrow v\ =\ sqrt{\dfrac{2 KQq_e}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2Kq_e\rho \times 2\pi R}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2\times 9\times 10^9\times 1.0\times 10^{-6}\times 2\times 3.14\times 0.7\times 1.6\times 10^{-19}}{9\times 10^{-31}\times \sqrt{0.7^2\ +\ 0.5^2}}}\\\Rightarrow v\ =\ 1.2\times 10^8\ m/s.$

Hence the velocity of the electron on the center of the charged ring is $1.2\times 10^8\ m/s.$

5 0

3 years ago

Low-frequency red light is used in dark rooms because it does not expose the film. This is because the red light has a low

Alex

energy. It will not disrupt the picture developing process by overexposing too much light on the film.

4 0

3 years ago

Read 2 more answers

To get a feeling for inertial forces discuss the familiar cases of accelerating in a car in a straight line while increasing or

inn [45]

Answer:

Explanation:

When we accelerate in a car on a straight path we tend to lean backward because our lower body part which is directly in contact with the seat of the car gets accelerated along with it but the upper the upper body experiences this force later on due to its own inertia. This force is accordance with Newton's second law of motion and is proportional to the rate of change of momentum of the upper body part.

Conversely we lean forward while the speed decreases and the same phenomenon happens in the opposite direction.

While changing direction in car the upper body remains in its position due to inertia but the lower body being firmly in contact with the car gets along in the direction of the car, seems that it makes the upper body lean in the opposite direction of the turn.

On abrupt change in the state of motion the force experienced is also intense in accordance with the Newton's second law of motion.

7 0

3 years ago

Gold-leaf electroscope uses

Kruka [31]

-identifies an electric charge
-it can identify its polarity (positive or negative) if you compare it to a charge that you already know
-can identify the magnitude of a charge (how big of a charge it is)

7 0

3 years ago

A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.1m facing each other,

borishaifa [10]

Answer:

8.854 pF

Explanation:

side of plate = 0.1 m ,

d = 1 cm = 0.01 m,

V = 5 kV = 5000 V

V' = 1 kV = 1000 V

Let K be the dielectric constant.

So, V' = V / K

K = V / V' = 5000 / 1000 = 5

C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F

C = 8.854 pF

5 0

3 years ago