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Salsk061 [2.6K]
3 years ago
7

Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi

eld and then injected into a region of constant magnetic field with a field strength of 0.65 T.
What is the potential difference in volts required in the first part of the experiment to accelerate electrons to a speed of 6 1 × 107 m/s?
Physics
1 answer:
yanalaym [24]3 years ago
8 0

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=v=6.1\times 10^7m/s

Charge on electron, q=e=1.6\times 10^{-19} C

Mass of electron,m_e=9.1\times 10^{-31} kg

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

V=\frac{v^2m_e}{2e}

Where V=Potential difference

m_e=Mass of electron

v=Velocity of electron

Using the formula

V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}

V=10581.59 V

Hence, the potential difference=10581.59 V

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A bulb converts 9J of energy in each second from an input of 100J of energy. Calculate efficiency of this light bulb
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3 years ago
7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in
viva [34]

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

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