
An electron moving with uniform velocity in x direction enters a region of uniform magnetic field along y direction. Which of the following physical quantity(ies) is (are) non-zero and remains constant?
I. Velocity of the electron.
II. Magnitude of the momentum of the electron.
III. Force on the electron.
IV. The kinetic energy of an electron.
A. only I and II
B. only III and IV
C. All four
D. only II and IV
Answer
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Hint: We will find the force acting on a charged particle due to magnetic field given by: ${{F}_{B}}=q(\overline{v}\times \overline{B})=qvB\sin \theta $. We will also use the knowledge of magnetic force acting upon a charged particle to be a deflecting force only. This means that the magnitude of the velocity wouldn’t change when passing through the magnetic field.
Step by step solution:
Let us understand how an uniform magnetic force along a certain direction acts on a charge travelling in the magnetic field.
As per the question, we are given that the magnetic field is along the y-direction. Further, the electron is said to be travelling along the x-direction with a velocity (v).
We know that magnetic force acting upon a charged particle is a deflecting force. It doesn’t lead to any increase or decrease in the velocity of the charged particle travelling in the magnetic field. Hence, the initial velocity of the charged particle upon entering the magnetic field remains the same throughout. Therefore, the velocity of the electron remains constant.
For the current case, the magnetic field is perpendicular to the velocity of the electron. The force acting on the electron due to the magnetic field is given by: ${{F}_{B}}=q(\overline{v}\times \overline{B})=qvB\sin \theta =qvB\sin {{90}^{0}}=qvB$. All the values in the magnetic force acting on the electron are constant. Hence, the force always remains constant.
Since, the force remains constant, therefore, no work is done by the magnetic field on the motion of the electron. This implies, the kinetic energy acting on the electron also remains constant throughout.
Since, the velocity of the electron remains constant throughout, the momentum (p) of the electron given by:\[\overline{~p}=m\overline{v}\] will remain constant throughout as well. Since, the mass of the particle under non-relativistic velocity conditions remains constant, hence the magnitude of the momentum of the electron given by:\[\left| \overline{p} \right|=\left| m\overline{v} \right|\] will remain constant throughout this process as well.
Therefore, all the four components remain constant, which is Option C.
Note:
The direction in which the electron starts to move due to the magnetic field can be found using Fleming's Left hand rule. The Index finger refers to the Intensity of magnetization, the next finger refers to the direction along which the charged particle is moving. The direction of the thumb shows the direction along which the charged particle gets deflected due to the perpendicular magnetic field.
Step by step solution:
Let us understand how an uniform magnetic force along a certain direction acts on a charge travelling in the magnetic field.
As per the question, we are given that the magnetic field is along the y-direction. Further, the electron is said to be travelling along the x-direction with a velocity (v).
We know that magnetic force acting upon a charged particle is a deflecting force. It doesn’t lead to any increase or decrease in the velocity of the charged particle travelling in the magnetic field. Hence, the initial velocity of the charged particle upon entering the magnetic field remains the same throughout. Therefore, the velocity of the electron remains constant.
For the current case, the magnetic field is perpendicular to the velocity of the electron. The force acting on the electron due to the magnetic field is given by: ${{F}_{B}}=q(\overline{v}\times \overline{B})=qvB\sin \theta =qvB\sin {{90}^{0}}=qvB$. All the values in the magnetic force acting on the electron are constant. Hence, the force always remains constant.
Since, the force remains constant, therefore, no work is done by the magnetic field on the motion of the electron. This implies, the kinetic energy acting on the electron also remains constant throughout.
Since, the velocity of the electron remains constant throughout, the momentum (p) of the electron given by:\[\overline{~p}=m\overline{v}\] will remain constant throughout as well. Since, the mass of the particle under non-relativistic velocity conditions remains constant, hence the magnitude of the momentum of the electron given by:\[\left| \overline{p} \right|=\left| m\overline{v} \right|\] will remain constant throughout this process as well.
Therefore, all the four components remain constant, which is Option C.
Note:
The direction in which the electron starts to move due to the magnetic field can be found using Fleming's Left hand rule. The Index finger refers to the Intensity of magnetization, the next finger refers to the direction along which the charged particle is moving. The direction of the thumb shows the direction along which the charged particle gets deflected due to the perpendicular magnetic field.
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