
An electron is travelling in east direction and a magnetic field is applied in upward direction then electron will deflect in
A. South
B. North
C. West
D. East
Answer
162.3k+ views
Hint: To find the direction of the magnetic force on the moving electron in the magnetic field, we can use either the Fleming’s left hand rule or the Lorentz’s force law. When we use Lorentz's force law we make equivalence of the geographic direction with the Cartesian coordinate system.
Formula used:
\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\], here \[\overrightarrow F \] is the force acting on the charged particle moving with velocity \[\overrightarrow v \] in a region of magnetic field \[\overrightarrow B \]
Complete step by step solution:
It is given that the electron is travelling in the east direction. Let the geographic direction towards east be equivalent to the +x-axis. Let the speed of the electron is v then the velocity vector will be \[\overrightarrow v = v\widehat i\]. The magnetic field is applied in the upward direction. The upward direction is equivalent to the +z-axis. Let the magnitude of the magnetic field is B then the magnetic field vector is \[\overrightarrow B = B\widehat k\]
Then the magnetic force on the electron is,
\[\overrightarrow F = - e\left( {v\widehat i \times B\widehat k} \right)\]
\[\Rightarrow \overrightarrow F = - evB\left( { - \widehat j} \right)\]
\[\therefore \overrightarrow F = evB\widehat j\]
The magnetic force on the electron is towards +y-axis. As the +y-axis of the Cartesian coordinate equivalent to the direction North. Hence, the magnetic force on the electron is acting on the North.
Therefore, the correct option is B.
Note: Using Fleming’s left hand rule, pointing the middle finger of left hand towards the magnetic field, i.e. towards upward and index finger towards the motion of the electron, i.e. east then the thumb will point towards the magnetic force, i.e. towards North.
Formula used:
\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\], here \[\overrightarrow F \] is the force acting on the charged particle moving with velocity \[\overrightarrow v \] in a region of magnetic field \[\overrightarrow B \]
Complete step by step solution:
It is given that the electron is travelling in the east direction. Let the geographic direction towards east be equivalent to the +x-axis. Let the speed of the electron is v then the velocity vector will be \[\overrightarrow v = v\widehat i\]. The magnetic field is applied in the upward direction. The upward direction is equivalent to the +z-axis. Let the magnitude of the magnetic field is B then the magnetic field vector is \[\overrightarrow B = B\widehat k\]
Then the magnetic force on the electron is,
\[\overrightarrow F = - e\left( {v\widehat i \times B\widehat k} \right)\]
\[\Rightarrow \overrightarrow F = - evB\left( { - \widehat j} \right)\]
\[\therefore \overrightarrow F = evB\widehat j\]
The magnetic force on the electron is towards +y-axis. As the +y-axis of the Cartesian coordinate equivalent to the direction North. Hence, the magnetic force on the electron is acting on the North.
Therefore, the correct option is B.
Note: Using Fleming’s left hand rule, pointing the middle finger of left hand towards the magnetic field, i.e. towards upward and index finger towards the motion of the electron, i.e. east then the thumb will point towards the magnetic force, i.e. towards North.
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