A positively charged (+q) particle of mass m | vedclass

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(B) The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.From this,the velocity $v$ is $v = \sqrt{\frac{2K}{m}}$.The magnetic force on a moving ch

Vedclass · Accepted Answer

$\frac{qB\sqrt{2K}}{m^{3/2}}$

Vedclass · Answer

(B) The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.From this,the velocity $v$ is $v = \sqrt{\frac{2K}{m}}$.The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the particle enters the magnetic field perpendicularly,the magnitude of the force is $F = qvB \sin(90^{\circ}) = qvB$.Substituting the expression for $v$,we get $F = qB \sqrt{\frac{2K}{m}}$.The acceleration $a$ is given by $a = \frac{F}{m}$.Therefore,$a = \frac{qB}{m} \sqrt{\frac{2K}{m}} = \frac{qB\sqrt{2K}}{m \cdot m^{1/2}} = \frac{qB\sqrt{2K}}{m^{3/2}}$.

$A$ positively charged $(+q)$ particle of mass $m$ has kinetic energy $K$ and enters a region with a horizontal magnetic field of induction $\vec{B}$ in a vertically downward direction. The acceleration of the particle is:

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