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(B) The particle is moving in a region with both electric field $\vec E = E\hat j$ and magnetic field $\vec B = B\hat j$. The force on the particle is

Vedclass · Accepted Answer

$\sqrt {{v^2} + \frac{{2qE}}{{m}}y}$

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(B) The particle is moving in a region with both electric field $\vec E = E\hat j$ and magnetic field $\vec B = B\hat j$. The force on the particle is $\vec F = q(\vec E + \vec v 	imes \vec B)$. The electric force acts only along the $y$-axis,causing acceleration $a_y = \frac{qE}{m}$. Using the equation of motion $v_y^2 = u_y^2 + 2a_y y$,where $u_y = 0$ (initial velocity has no $y$-component),we get $v_y^2 = 2 \left( \frac{qE}{m} ight) y$. The magnetic force $\vec F_m = q(\vec v 	imes \vec B)$ is always perpendicular to the velocity,so it does no work and does not change the speed of the particle. The initial speed is $v_0 = |\vec v| = v$. The final speed $v_f$ is given by $v_f = \sqrt{v_x^2 + v_y^2 + v_z^2}$. Since the magnetic force does no work,the change in kinetic energy is due only to the electric field: $\frac{1}{2}mv_f^2 - \frac{1}{2}mv^2 = qEy$. Thus,$v_f^2 = v^2 + \frac{2qEy}{m}$,which gives $v_f = \sqrt{v^2 + \frac{2qEy}{m}}$.

$A$ positive charge particle having charge $q$ and mass $m$ has velocity $\vec v = v\left( {\frac{{\hat i + \hat k}}{{\sqrt 2 }}} \right)$ in the magnetic field $\vec B = B\hat j$ and electric field $\vec E = E\hat j$ at the origin. Its speed as a function of $y$ is:

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