1 Physics -- Electrons

Write down expression of acceleration of a moving charge Q in a parallel and perpendicular magnetic field.

1 Answers

The expression for the acceleration of a moving charge Q in a magnetic field depends on the direction of the magnetic field relative to the velocity of the charge. Here are the expressions for the acceleration of a moving charge Q in a parallel and perpendicular magnetic field:

1. Parallel Magnetic Field:

If the magnetic field is parallel to the velocity of the charge, the charge experiences a force that causes a centripetal acceleration. The expression for the acceleration (a_parallel) is given by:

a_parallel = (Q * B * v) / m

where:- a_parallel is the acceleration of the charge (in m/s²)

- Q is the charge of the particle (in coulombs)

- B is the magnetic field strength (in teslas)

- v is the velocity of the charge (in m/s)

- m is the mass of the charge (in kilograms)

2. Perpendicular Magnetic Field:

If the magnetic field is perpendicular to the velocity of the charge, the charge experiences a force that causes it to move in a circular path. The expression for the acceleration (a_perpendicular) is given by:

a_perpendicular = (Q * B * v) / m

where:- a_perpendicular is the acceleration of the charge (in m/s²)

- Q is the charge of the particle (in coulombs)

- B is the magnetic field strength (in teslas)

- v is the velocity of the charge (in m/s)

- m is the mass of the charge (in kilograms)

In both cases, the direction of the acceleration is perpendicular to both the magnetic field and the velocity of the charge, as dictated by the right-hand rule.

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I need help

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At very low pressure, the density of the gas in the discharge tube becomes very low. As a result, the collision between the electrons emitted by the cathode and the gas molecules are avoided. Due to this, the gas molecules do not get excited. Since the gas molecules are not excited, there are no transitions of the molecules from the excited states to the ground states and hence no light is emitted. Therefore, the discharge tube at very low pressure appears dark.

1. Can we perform Millikan's experiment with drop of any size?

No, it is not advised to perform Millikan's oil drop experiment with drops of any size due to limitations of experimental accuracy.

When we use a drop of larger size (as compared to microscopic sizes), they will have comparatively higher velocity which is difficult to detect.

Similarly, a large electric field should be applied to move the drop towards the positive terminal against gravity.

Why specific charge for positive rays much smaller that that of cathode rays?

Specific charge is the ratio of a particle's charge to its mass.

Since, the mass of an electron is about 1/1837 times that of the mass of a proton. Its mass is much smaller than the proton.

A positive ray contains alpha particles (doubly ionized He nucleus). Its mass is much larger than that of an electron so, the greater electronic charge has negligible effect as compared to its mass, and the specific charge is much smaller than that of the electrons.

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Following are the significance of Millikan's oil drop experiment:

It verifies that electronic charge is the smallest possible unit of charge.
It helps to calculate the electronic charge by using the value of the specific charge of electron (e/m).

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Any two properties of cathode rays are:

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They travel with 1/10 of the velocity of light.

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