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A particle having mass m and charge q is released from the origin in a region in which electric field and magnetic field is given by $\vec B = - {B_o}\hat j$and$\vec E = {E_o}\hat k$. Find the speed of the particle as a function of its z-coordinate.
A) $\sqrt {\dfrac{{qEz}}{m}} $
B) $\sqrt {\dfrac{{2\left( {qvB + qE} \right)z}}{m}} $
C) $\sqrt {\dfrac{{\left( { - qvB + qE} \right)2z}}{m}} $
D) $\sqrt {\dfrac{{2qEz}}{m}} $

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Last updated date: 20th Jun 2024
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Answer
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Hint:The charge particle will have force due to the electric field and due to the magnetic field. The particle will move in the direction resultant to the force exerted by the force on the particle due to electric and magnetic fields.

Formula used:The formula of the force due to magnetic field is given ${\vec F_m} = q \cdot \left( {\vec v \times \vec B} \right)$ where q is the charged particle v is the velocity and B is the magnetic field. The force due to the electric field on the charged particle is given by ${\vec F_e} = q \cdot \vec E$ where q is the charged particle and E is the electric field.

Step by step solution:
The total force on the charged particle is given by
$ \Rightarrow \vec F = {\vec F_m} + {\vec F_e}$
The force F is the total force and ${F_m}$ is the force due to the magnetic field and ${F_e}$ is the force due to the electric field.
$ \Rightarrow \vec F = q \cdot \left( {\vec v \times \vec B} \right) + q \cdot \vec E$
$ \Rightarrow \vec F = q \cdot \left[ {\left( {\vec v \times \vec B} \right) + \vec E} \right]$.........eq. (1)
Let the initial velocity is given by,
$u = {u_x}\hat i + {u_y}\hat j + {u_z}\hat k$.
Replacing the value of the velocity in the equation (1)
$ \Rightarrow \vec F = q \cdot \left[ {\left( {\vec v \times \vec B} \right) + \vec E} \right]$
$ \Rightarrow \vec F = q \cdot \left[ {\left( {{u_x}\hat i + {u_y}\hat j + {u_z}\hat k} \right) \times \left( { - {B_o}\hat j} \right) + {E_o}\hat k} \right]$
Applying cross product.
$ \Rightarrow \vec F = q \cdot \left[ {\left( { - {u_x}{B_o}\hat k + {u_z}{B_o}\hat i} \right) + {E_o}\hat k} \right]$
Since the velocity in the x-direction is zero as the particle is released at origin. Therefore the force in the z-direction will be.
$ \Rightarrow {\vec F_z} = q \cdot \left( {{E_o}\hat k} \right)$
Since force is given by $F = ma$ ,the acceleration will be equal to.
$ \Rightarrow a = \dfrac{F}{m}$
$ \Rightarrow {a_z} = \dfrac{{q{E_o}}}{m}$………eq. (1)
According to the Newton’s law of motion,
$ \Rightarrow {v^2} - {u^2} = 2as$
Replace the value of${a_z}$ we get
$ \Rightarrow {v^2} = 2\left( {\dfrac{{q{E_o}}}{m}} \right)z$
$ \Rightarrow v = \sqrt {2\left( {\dfrac{{q{E_o}}}{m}} \right)z} $.
The velocity of the charged particle in the z-direction is given by$v = \sqrt {2\left( {\dfrac{{q{E_o}}}{m}} \right)z} $.

The correct option for this problem is option D

Note:The acceleration in the z-direction is asked in the problem. The acceleration is constant because of which we are able to apply Newton's law of motion equations, if the acceleration is not constant we cannot apply the equations of Newton's law.