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A proton moving along the positive x-axis enters a uniform magnetic field which is directed along the

SSSSS [86.1K]

Answer: The force acting on the proton of charge 1.5 x10-19 C moving with velocity 1.5 x10-19 C under the influence of a magnetic field of 0.7 T will be 3.15×10^-13 N.

Explanation: To find the answer we need to know more about the Lorentz magnetic force.

<h3>What is the Lorentz magnetic force acting on the proton?</h3>

Consider a proton of charge q moving with a velocity v in a magnetic field, then the Lorentz magnetic force exerted on the proton can be expressed as,

F= q (v× B)

$F= qvBsin\alpha$ where, $\alpha$ is the angle between v and B.

In the question, it is given that,

$B=0.7 T\\q=1.5*10^{-19}C\\v= 3*10^{6}m/s.\\\alpha =90 degree.\\$ because, from the question it is clear that the proton is moving along x axis and the magnetic field is along the y axis.

Thus, we can find the force acting on the proton as,

$F=qvBsin\alpha =1.5*10^{-19}C*3*10^6 m/s*0.7T*sin (90)\\F=3.15*10^{-13}N$

Thus, we can conclude that the Lorentz force acting on the proton will be 3.15×10^-13 N.

Learn more about the Lorentz magnetic force here:

brainly.com/question/28047923

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6 0

1 year ago

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When a permanent magnet picks up an iron nail, the nail also becomes a magnet. How does the magnetized nail differ from an elect

Yuri [45]

Hi u need help still?

4 0

2 years ago

An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)

weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

$\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767$

T₂ = 290.15 × 3.17767 = 922.00139

$\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5$

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

$\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3} = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12$

$\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702$

Therefore;

$T_4 = \frac{1383.002}{2.702} =511.859 \ k$

$Q_1 = c_p(T_3-T_2)$

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

$c_v(T_4-T_1)$ = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

$\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1 \right )r_{v}^{k-1}}$

Where:

β = Cut off ratio

Plugging in the values, we get;

$\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1 \right )18^{1.4-1}}= 0.5191$

Therefore;

$\eta = \frac{\sum Q}{Q_1}$

$\therefore 0.5191 = \frac{150}{Q_1}$

Heat supplied = $\frac{150}{0.5191} = 288.978 \ hp$

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0

3 years ago

A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?

olchik [2.2K]

Answer:

$m=5.78\times 10^{-3}\ g$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Explanation:

Given:

quantity of charge transferred, $Q=31\ C$

<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron $1.602\times 10^{-19}\ C$

so,

$n_e=\frac{Q}{e}$

$n_e=\frac{31}{1.602\times 10^{-19}}$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Now if each water molecules donates one electron:

Then we require $n=1.935\times 10^{20}$ molecules.

<u>Now the no. of moles in this many molecules:</u>

$n_m=\frac{n}{N_A}$

where

$N_A=6.022\times 10^{23}$ Avogadro No.

$n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}$

$n_m=3.213\times 10^{-4}\ moles$

We have molecular mass of water as M=18 g/mol.

<u>So, the mass of water in the obtained moles:</u>

$n_m=\frac{m}{M}$

where:

m = mass in gram

$3.213\times 10^{-4}=\frac{m}{18}$

$m=5.78\times 10^{-3}\ g$

7 0

2 years ago

This is science btw please help me

raketka [301]

Data charts would use descriptive statistics to show accurate reading measured throughout the lab. Charts and graphs can also be used to show the progress and result of the lab.

7 0

2 years ago

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