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lutik1710 [3]
3 years ago
8

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec

tor. Ignoring the particle's weight, what type of path will the particle follow
Physics
1 answer:
Ugo [173]3 years ago
3 0

Answer:

a circular path

Explanation:

In a magnetism field if a charged particle having a charge of magnitude '' enters  such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F

F = V\times B

According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.

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3 years ago
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Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7
sleet_krkn [62]

Answer : \underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Explanation :

Given that,

Charge of particle 1 =  -1.50\times10^{-7} c

Distance x = 6 cm

Charge of particle 2 = 1.50\times10^{-7} c

Distance x = 27 cm

Total distance = \dfrac{6+27}{2}

r = 16.5\ cm

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm

Formula of electric field

E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}

Now, the the electric field due to  particle 1

\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i}  \dfrac{N}{C}

\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i}  \dfrac{N}{C}

\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Similarly, the electric field due to particle 2

\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Resultant Electric field

\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}

\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Hence, this is the required answer.






3 0
3 years ago
What average force is needed to accelerate a 7.00-gram pellet from rest to 155 m/s over a distance of 0.600 m along the barrel o
leonid [27]

Answer: The force needed is 140.22 Newtons.

Explanation:

The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.

The force exerted on the pellet is

F = m a

In order to calculate the acceleration, given the displacement d,  

d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}

This value can be used to determine the time for the pellet through the barrel:

t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s

Finally, we can use the above to calculate the force:

F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N



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Answer:

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Explanation:

descriptive, because scientists are writing down the observations but not making comparisons.  

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What is Intracellular communication
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Answer:

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