A particle of mass M and positive charge Q,mo | vedclass

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(C) The initial velocity is $\vec{u}_1 = 4\hat{i} 	ext{ m/s}$. The final velocity is $\vec{u}_2 = 2\sqrt{3}\hat{i} + 2\hat{j} 	ext{ m/s}$. Since the

Vedclass · Accepted Answer

$(A, C)$

Vedclass · Answer

(C) The initial velocity is $\vec{u}_1 = 4\hat{i} 	ext{ m/s}$. The final velocity is $\vec{u}_2 = 2\sqrt{3}\hat{i} + 2\hat{j} 	ext{ m/s}$. Since the magnetic force is perpendicular to the velocity,the speed remains constant: $|\vec{u}_1| = |\vec{u}_2| = 4 	ext{ m/s}$. The angle of deviation $	heta$ is given by $	an 	heta = \frac{v_y}{v_x} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,so $	heta = 30^\circ = \frac{\pi}{6} 	ext{ rad}$. The particle deflects in the $+y$ direction,implying the magnetic force $\vec{F} = Q(\vec{v} 	imes \vec{B})$ has a positive $y$-component. Given $\vec{v}$ is in the $x$-direction,$\hat{i} 	imes \vec{B}$ must have a positive $j$-component,which occurs if $\vec{B}$ is in the $-z$ direction. The time taken to traverse the arc is $t = \frac{	heta}{\omega} = \frac{	heta M}{QB}$. Given $t = 10 	ext{ ms} = 0.01 	ext{ s}$,we have $0.01 = \frac{\pi/6 \cdot M}{QB}$. Solving for $B$: $B = \frac{\pi M}{6 \cdot 0.01 \cdot Q} = \frac{100\pi M}{6Q} = \frac{50\pi M}{3Q}$. Thus,statements $(A)$ and $(C)$ are correct.

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