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neonofarm [45]
3 years ago
6

An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi

eld with x component 0.027 T and ycomponent -0.15 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.
I know that the equation is FB = qv × B, I'm just having difficulty computing the cross product directly? Also, I know that part b is the same answer as part a just in the opposite direction
Physics
1 answer:
likoan [24]3 years ago
6 0

Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force,  a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as:  \overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\

In respect to the magnetic field; \overrightarrow{B}=(0.027 i - 0.15 j) [T]

The magnetic force can be written as;

\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]

Bear in mind q =-1.6021x10^{-19} [C]  

thus,

\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, k^{2}=k\cdot k = 1

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, k^{2}=k\cdot k = 1

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.  

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The image height is 3.0 cm

Explanation:

Given;

object distance, d_o = 15.0 cm

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What is the result of (305.120 + 267.443) x 0.50? How many answers can be written based on the principle of significant digits?
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(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

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For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

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<span>
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