) An electron moving along the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, what is the direction of the magnetic field in this region

Answers

Answer 1
Answer:

Answer:

- z direction

Explanation:

To find the direction of the magnetic field, you take into account that the magnetic force over a charge, is given by the following cross product:

\vec{F_B}=q\vec{v}\ X\ \vec{B}      (1)

F_B: magnetic force

q: charge of the particle

v: velocity of the charge

B: magnetic field

In this case you have that the electron is moving along x-axis. You can consider this direction as the ^i direction. The electron experiences a magnetic deflection in the -y direction, that is, in the -^j  direction.

By the cross products between unit vectors, you have that:

-^j = ^i X ^k

That is, the cross product between two vectors, one in the +x direction, and another one in the +z direction, generates a vector in the -y direction. However, it is necessary to take into account that the negative charge of the electron change the sign of the result of the cross product, which demands that the second vector is in the -z direction. That is:

-^i X -k^ = ^i X ^k = - ^j

Hence, the direction of the magnetic field is in the -z direction


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If the rise and fall of your lungs is considered to be simple harmonic motion, how would you relate the period of the motion to your breathing rate (breaths per minute)? Breaths per minute is an angular frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is its reciprocal. Breaths per minute is an angular frequency. The period is its reciprocal.

Answers

Answer:

Breaths per minute is a frequency. The period is its reciprocal.

Explanation:

In simple harmonic motion, a period (T) is the time taken for one point to start in a position and reach that position again, in other words to complete a cycle or lapse. In this case, a period is the time one takes from starting to inspire the air to releasing all of it from the lungs.

In simple harmonic motion, the frequency (f) is how many times a point completes a cycle or lapse in one unity of time (could be one second, one minute, one hour, etc). In this case, the frequency is how many times one breathes in one minute. This is the breathing rate, since it is breathings per minute. Breaths per minute is a frequency.

Period (T) and frequency (f) relate to each other in the following formulae: T=(1)/(f) or f=(1)/(T) .

Therefore, breaths per minute is a frequency, and since it is related to the period, we say the period is reciprocal to it.

An object essentially at infinity is moved to a distance of 90 cm in front of a thin positive lens. In the process its image distance triples. Determine the focal length of the lens.

Answers

Answer:

67.5 cm

Explanation:

u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm

let f be the focal length

Use lens equation

1 / f = 1 / v - 1 / u

1 / f = 1 / 270 + 1 / 90

1 / f = 4 / 270

f = 67.5 cm

Final answer:

To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.

Explanation:

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance.

Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:

1/f = 1/(3v) - 1/(90)

Multiplying through by 90*3v, we get:

90*3v/f = 270v - 90*3v

90*3v/f = 270v - 270v

90*3v/f = 0

Simplifying further, we find that: v = 0

When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.

Ultrasonic imaging is made possible due to the fact that a sound wave is partially reflected whenever it hits a boundary between two materials with different densities within the body. the percentage of the wave reflected when traveling from material 1 into material 2 is r=(ρ1−ρ2ρ1+ρ2)2. knowing this, why does the technician apply ultrasound gel to the patient before beginning the examination?

Answers

Answer:

The gel that is applied before ultrasonic imaging is a conducting material. It acts as a medium between transducer and skin. The ultrasonic waves easily transmit from the probe to the tissues because of gel. A tight bond is created between the probe and skin layer and the gel acts as a coupling agent. The density of the gel is similar to the skin layer. This reduces the attenuation of the waves.  A thin layer of gel is applied which fills the air gaps and helps in transmission of waves to the tissues. Hence, the technician apply ultrasound gel to the patient before beginning the examination

The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection.


A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can reach.

Answers

Answer:

h = 16.9 m

Explanation:

When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

(0.5)m(Vf² - Vi²) = mgh

h = (0.5)(Vf² - Vi²)/g

where,

Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

Vi = Initial Speed of Ball = 18.2 m/s

g = acceleration due to gravity = - 9.8 m/s² ( negative for upward motion)

h = maximum height the ball can reach = ?

Therefore, using values in the equation, we get:

h = (0.5)[(0 m/s)² - (18.2 m/s)²]/(-9.8 m/s²)

h = 16.9 m

The electric field just above the surface of the charged drum of a photocopying machine has a magnitude E of 2.5 × 105 N/C. What is the surface charge density on the drum, assuming that the drum is a conductor?

Answers

Answer:

Charge_(density)=2.2125*10^(-6)C/m^(2)

Explanation:

Given data

Electric Field E=2.5×10⁵ N/C

To find

Charge Density

Solution

From definition of charge density we know that:

Charge Density=Electric field×Permttivity

Where Permttivity ∈₀=8.85×10⁻¹²C²/N.m²

Charge_(density)=(2.5*10^(5)N/C)*(8.85*10^(-12)C^(2)/N.m^(2))\n Charge_(density)=2.2125*10^(-6)C/m^(2)

What sound frequency could a human detect

Answers

Answer:

People can hear sounds at frequencies from about 20 Hz to 20,000 Hz,

20 Hz up to 20,000 Hz

Brain pls