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Georgia [21]
3 years ago
5

A proton and an electron enter perpendicular to the direction of the magnetic field. The speed of the proton is twice the speed

of the electron. What is the ratio of the force experienced by the proton to the electron?
Physics
1 answer:
Nezavi [6.7K]3 years ago
3 0

Answer:

The force on the proton will be twice in comparison with the force experienced by the electron.

Explanation:

The magnetic force acting on the moving charge particle is perpendicular to the velocity as well as the magnetic field. The factors upon which the magnitude of the force is depending and proportional are as follows:

1 Magnitude of the charge particle.

2 Magnitude of the velocity of moving charge.

3 Magnitude of magnetic field.

4 Sine of angle between the velocity and magnetic field.

As far as direction is concerned, it can be find out by right hand rule.

Lets consider,

Speed of electron = Ve

Speed of proton vp= 2ve

Magnitude of the force F = qvBsin∅

Force acting  on electron Fe = qe.ve.Bsin∅

Force acting on proton Fp = qp.vp.Bsin∅

                                       Fp = -2 qe.ve.Bsin∅

                                       Fp = -2Fe

Negative sign shows the direction of Force on proton is opposite to the direction of force on electron.

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A student places his hand in front of a plane mirror as shown in
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Answer:

A:

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7 0
4 years ago
The earth exerts a gravitational force of 850N on John. What is Johns mass in kg
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Answer:

86.73 kg

Explanation:

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850N=m9.8m/s^2

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5 0
2 years ago
2. Is the original mixture homogeneous or heterogeneous? Why? .
myrzilka [38]

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4 0
3 years ago
slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0
3 years ago
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