Physics, 12.03.2020 06:18 michibabiee

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It enters a magnetic field of ( ) mT. What is the acceleration of the particle?

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Answer from: owlsandR5

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The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

F = ma

Where F is the force experience by a particle

m is the mass of the particle

a is the acceleration of the particle

And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

F = qvB

Where q is the charge of the particle

v is the velocity of the charge

B is the magnetic field the charge is under it influence

Now equating this two formulas

ma = qvB

Making a the subject we have

$a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is hence the direction would be in the direction

So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

$a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

$a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

a = 0.003 * 3i(2i+3j+4k)

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

$i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

$a = 0.003* 6 =0.018\ m/s^2$

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a

Answer from: Quest

Answerthe force between the charges will be f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges: $f = \dfrac{k q_1 q_2}{r^2}$ k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then $\begin{aligned} f & = \dfrac{k q_1 q_2}{r^2} \\ f_{3r} & = \dfrac{k q_1 q_2}{(3r)^2} \\ & = \dfrac{k q_1 q_2}{3^2 \cdot r^2} \\ & = \frac{1}{9} \cdot \boxed{\dfrac{k q_1 q_2}{r^2} } \\ & = \frac{1}{9} f & & \left(\text{\footnotesize since $ f = \tfrac{k q_1 q_2}{r^2}$}\right) \end{aligned}$

Answer from: Quest

One answer could be erosion

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