Two charged particles A and B,each of charge | vedclass

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(C) In a velocity selector,the electric force and magnetic force must balance for the particle to pass undeflected: $qE = qvB$,so $E = vB$.Given $v =

Vedclass · Accepted Answer

The ratio $r_A / r_B$ of the radii of the circular paths for the two particles is $12/13$.

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(C) In a velocity selector,the electric force and magnetic force must balance for the particle to pass undeflected: $qE = qvB$,so $E = vB$.Given $v = 1.5 	imes 10^6 \, ms^{-1}$ and $B = 1.0 \, T$,we have $E = (1.5 	imes 10^6)(1.0) = 1.5 	imes 10^6 \, Vm^{-1}$.Since the magnetic force $\vec{F}_m = q(\vec{v} 	imes \vec{B})$ acts in the $+x$ direction (using right-hand rule for $\vec{v} = v\hat{j}$ and $\vec{B} = -B\hat{k}$),the electric force must act in the $-x$ direction. Thus,$\vec{E} = -1.5 	imes 10^6 \hat{i} \, NC^{-1}$.For the circular motion in chamber $X$,the radius is given by $r = \frac{mv}{qB}$.Since $v, q,$ and $B$ are the same for both particles,$r \propto m$.Therefore,$\frac{r_A}{r_B} = \frac{m_A}{m_B} = \frac{12}{13}$.Thus,option $C$ is correct.

List-$I$	List-$II$
$(A)$ Permeability of free space	$(I) \ [M L^2 T^{-2}]$
$(B)$ Magnetic field	$(II) \ [M T^{-2} A^{-1}]$
$(C)$ Magnetic moment	$(III) \ [M L T^{-2} A^{-2}]$
$(D)$ Torsional constant	$(IV) \ [L^2 A]$

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