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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The kinetic energy of the particle is given by $K = \frac{1}{2} m v^2$.From this,the velocity $v$ is $v = \sqrt{\frac{2 K}{m}}$.The magnetic force on a moving charge is given by $\overrightarrow{F} = q(\overrightarrow{v} 	imes \overrightarrow{B})$.The magnitude of the force is $F = q v B \sin 	heta$.Since the particle enters the field vertically downward and the field is horizontal,the angle $	heta = 90^{\circ}$.Therefore,$F = q v B = q B \sqrt{\frac{2 K}{m}}$.The acceleration $a$ is given by $a = \frac{F}{m}$.Substituting the value of $F$,we get $a = \frac{q B \sqrt{2 K}}{m \sqrt{m}} = \frac{q B \sqrt{2 K}}{m^{3/2}}$.

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

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