Ask question

Ask question

All categories

3 years ago

8

A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

+ 2ĵ − k) T. What is the magnitude of the magnetic force this particle experiences?

2 answers:

gavmur [86]3 years ago

8 0

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

(v × B) = (4î + 5ĵ + 14k)

F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

marishachu [46]3 years ago

5 0

Answer:

1.442×10^-18N

Explanation:

Magnitude of magnetic force experienced by the particle id expressed as;

F = qvB sin theta where;

q is the charge on the proton = 1.602×10^-19C

v is the velocity =(4î − 6ĵ + k) m/s

B is the magnetic field = (î + 2ĵ − k) Tesla

theta is the angle that the conductor make with the magnetic field = 90°

Substituting this values in the formula to get the magnitude of the force we have;

F = 1.602×10^-19 × (4î − 6ĵ + k)(î + 2ĵ − k) sin90°

Multiplying the vector function

(4î − 6ĵ + k)(î + 2ĵ − k) = 4i.i - 12j.j - k.k

Since i.i = j.j = k.k = 1

(4î − 6ĵ + k)(î + 2ĵ − k) = 4-12-1 = -9

F = 1.602×10^-19×(-9)×1

F = -14.418×10^-19N

F = -1.442 × 10^-18N

|F| = 1.442 × 10^-18N

The magnitude of the magnetic force this particle experiences is therefore 1.442 × 10^-18N

You might be interested in

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

6 0

3 years ago

In major league baseball, the pitcher's mound is 60 feet from the batter. If a pitcher throws a 85 mph fastball, how much time e

Nostrana [21]

The given from your problem are the following:
V = 85mph (This is miles per hour)
d = 60 feet

If you notice the units do not match. Before we can do anything else, we need to make the figures match.

In this case, we will convert 85miles per hour to feet per hour. There are 5,280 feet in 1 mile.
$\frac{85miles }{hr}$ x $\frac{5,280feet}{1miles} = \frac{448,800feet}{hr}$

But wait! If you think about the scenario, you are looking for how long it will take for the ball to reach the bat. The most applicable unit of time to use here is second. It would be very hard to really measure a short and instantaneous event in hours. So we convert it into feet per second:

There are 3,600 seconds in 1 hour.

$\frac{448,800feet }{hour}$ x $\frac{1hour}{3,600seconds} = \frac{448,800feet}{3,600 seconds} = 124.67ft/s$

So now we have our new given as:

v = 124.67ft/s
d = 60 ft

The formula for time can be derived from the formula from velocity, which is:
$velocity = \frac{distance}{time}$

The formula of time will then be:
$time= \frac{distance}{velocity}$

All you need to do is plug in what you know and solve for what you don't know.

$time= \frac{60feet}{124.67ft/s}$

$time= 0.48s$

The answer then is 0.48s.

If you want this in hours, just divide the value in seconds by 3,600. The answer would then be 0.00013hr. (See how small it is? This is why seconds would be a more appropriate measure.)

8 0

3 years ago

How many watt hours will 3-155amp hour 12 volt batteries wired in a parallel configuration produce?

Ede4ka [16]

Answer:

B 5580 W•hr

Explanation:

A Watt is a Volt times an Amp

3(12 V(155 A•hr)) = 5580 W•hr

4 0

3 years ago

What is the top of a wave called?

ohaa [14]

Answer:

The highest part of the wave is called the crest.

Explanation:) hope this helps

The highest part of the wave is called the crest. The lowest part is called the trough. The wave height is the overall vertical change in height between the crest and the trough and distance between two successive crests (or troughs) is the length of the wave or wavelength.

7 0

1 year ago

Which units are used to measure both velocity and speed? Select three options.

xz_007 [3.2K]

M/s, km/h, and mph are all used to measure these quantities

6 0

3 years ago

Read 2 more answers