Question

When a proton is released from the rest in a room, it starts with an initial acceleration ${{a}_{0}}$ towards north with a speed ${{v}_{0}}$ it moves with an initial acceleration 3${{a}_{0}}$ towards west. The electric and magnetic fields in the room are
A. $\dfrac{m{{a}_{0}}}{e}$west, $\dfrac{2m{{a}_{0}}}{e{{v}_{0}}}$up
B. $\dfrac{m{{a}_{0}}}{e}$west, $\dfrac{2m{{a}_{0}}}{e{{v}_{0}}}$down
C. $\dfrac{m{{a}_{0}}}{e}$west, $\dfrac{3m{{a}_{0}}}{e{{v}_{0}}}$up
D. $\dfrac{m{{a}_{0}}}{e}$east, $\dfrac{3m{{a}_{0}}}{e{{v}_{0}}}$down

Answer 1

The direction lines of electric field lines can be described as the direction in which a proton is accelerated when subjected to that electric field. The formula for both magnetic and electric forces have been given below. We also know that total force on any body is the product of its mass and its acceleration.

Formula used: $F=qE$
Where q is charge and E is electric field.
$F=q{{v}_{0}}B$
where B is the magnetic field.

Complete step by step answer:
Electric current or electric flow is the flow of electrons. Its direction is the direction of flow of electrons or opposite to the direction of flow of protons or positive charge. The direction lines of electric field lines can be described as the direction in which a proton is accelerated when subjected to that electric field.
Any charge experiences a force when put into a electric field this force is given by
$F=qE$
Initial acceleration of the proton is given as ${{a}_{0}}$ and the direction is given as west.
Magnitude of the charge on the proton is the same as the charge on the electron and is given as e. Let the mass of proton be m
Then, as proton is accelerated the force it experiences must be
${{F}_{1}}=m{{a}_{0}}$
Now this force is a result of Electric field

& m{{a}_{0}}=qE \\\ & \Rightarrow E=\dfrac{m{{a}_{0}}}{q} \\\ \end{aligned}$$ And the direction of this field is the same as the acceleration experienced by the proton, i.e. westward. Now, the proton has a velocity $${{v}_{0}}$$ and it has an acceleration of $${{v}_{0}}$$. Let the magnetic force the proton experience be $${{F}_{2}}$$ then $$\begin{aligned} & {{F}_{2}}+{{F}_{1}}=3m{{a}_{0}} \\\ & \Rightarrow {{F}_{2}}=2m{{a}_{0}} \\\ \end{aligned}$$ Also $$\begin{aligned} & {{F}_{2}}=q{{v}_{0}}B \\\ & \Rightarrow B=\dfrac{2m{{a}_{0}}}{q{{v}_{0}}} \\\ \end{aligned}$$ Since velocity is along north, force is along west, so the magnetic field is downwards. **So, the correct answer is “Option B”.** **Note:** Forces like Gravitation, Electrostatic and magnetic forces are called non-contact forces. There is direct contact needed for the body to experience these forces. These forces act on the body through their field lines. Any mass/charge place in their field experiences these forces.