Answer
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Hint:In this question,we are going to apply the concepts of electric field and magnetic field.In the velocity selector, the electric force acting on the ions is equal to the magnetic forces on them.
Complete step by step answer:
The velocity selector consists of an evacuated region of crossed electric and magnetic fields in the plane of the paper and normal to it. So, the electric field is parallel to the motion of the particles.
The orientations of the two-fields ensure the electric and magnetic forces on the ions are in the same plane and they are so adjusted that the deflections they produce on the ions are in exact opposition. The ions are thus able to pass through the selector without suffering any deflection. Thus, the electric force acting on the ions is equal to the magnetic forces on them.
The electric field in the velocity selector is \[\overrightarrow{E}\] and the magnetic field is \[\overrightarrow{B}\]. Only those ions of charge q will pass undeviated through the selector whose velocity v satisfies the relation
\[qE=Bqv\Rightarrow v=\dfrac{E}{B}\] or \[E=vB\]
So, option A is the correct answer.
Additional information:
The radius r of the ion-path in the field \[\overrightarrow{B'}\] is given by,
\[ B'qv=\dfrac{M{{v}^{2}}}{r}\], where \[M\] is the mass of the ion.
The velocity selector has direction focussing property and its addition to the mass spectrograph greatly improved its resolving power.
Note:The passage shows a Bainbridge mass spectrograph, which uses a velocity selector that makes the stream of positive ions perfectly homogenous before its entry into the magnetic field, no matter what the initial velocities of the ions are. The emergent ion has very little velocity-spread.
Complete step by step answer:
The velocity selector consists of an evacuated region of crossed electric and magnetic fields in the plane of the paper and normal to it. So, the electric field is parallel to the motion of the particles.
The orientations of the two-fields ensure the electric and magnetic forces on the ions are in the same plane and they are so adjusted that the deflections they produce on the ions are in exact opposition. The ions are thus able to pass through the selector without suffering any deflection. Thus, the electric force acting on the ions is equal to the magnetic forces on them.
The electric field in the velocity selector is \[\overrightarrow{E}\] and the magnetic field is \[\overrightarrow{B}\]. Only those ions of charge q will pass undeviated through the selector whose velocity v satisfies the relation
\[qE=Bqv\Rightarrow v=\dfrac{E}{B}\] or \[E=vB\]
So, option A is the correct answer.
Additional information:
The radius r of the ion-path in the field \[\overrightarrow{B'}\] is given by,
\[ B'qv=\dfrac{M{{v}^{2}}}{r}\], where \[M\] is the mass of the ion.
The velocity selector has direction focussing property and its addition to the mass spectrograph greatly improved its resolving power.
Note:The passage shows a Bainbridge mass spectrograph, which uses a velocity selector that makes the stream of positive ions perfectly homogenous before its entry into the magnetic field, no matter what the initial velocities of the ions are. The emergent ion has very little velocity-spread.
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