At t = 0,a positively charged particle of mas | vedclass

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(D) The initial velocity is $\vec{u} = u_0 \cos(37^o) \hat{i} + u_0 \sin(37^o) \hat{j} = u_0 (\frac{4}{5} \hat{i} + \frac{3}{5} \hat{j})$.Since the ma

Vedclass · Accepted Answer

$u_0 \left[ \frac{\sqrt{39}}{10} \hat{i} + \frac{3}{5} \hat{j} + \frac{1}{2} \hat{k} \right]$

Vedclass · Answer

(D) The initial velocity is $\vec{u} = u_0 \cos(37^o) \hat{i} + u_0 \sin(37^o) \hat{j} = u_0 (\frac{4}{5} \hat{i} + \frac{3}{5} \hat{j})$.Since the magnetic field is $\vec{B} = B_0 \hat{j}$,the force on the particle is $\vec{F} = q(\vec{v} 	imes \vec{B})$.This force is always perpendicular to the magnetic field,so the component of velocity along the $y$-axis $(v_y)$ remains constant. Thus,$v_y = u_0 \sin(37^o) = \frac{3}{5} u_0$.Also,the speed of the particle remains constant in a magnetic field,so $|\vec{v}|^2 = u_0^2 = v_x^2 + v_y^2 + v_z^2$.Substituting $v_y = \frac{3}{5} u_0$,we get $v_x^2 + v_z^2 = u_0^2 - (\frac{3}{5} u_0)^2 = \frac{16}{25} u_0^2$.Checking the options,for option $(D)$: $v_x^2 + v_z^2 = u_0^2 [(\frac{\sqrt{39}}{10})^2 + (\frac{1}{2})^2] = u_0^2 [\frac{39}{100} + \frac{1}{4}] = u_0^2 [\frac{39+25}{100}] = u_0^2 [\frac{64}{100}] = \frac{16}{25} u_0^2$. This matches the condition.

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