(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 =

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

(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 $v = 1.5 \times 10^6 \, ms^{-1}$ and $B = 1.0 \, T$,we have $E = (1.5 \times 10^6)(1.0) = 1.5 \times 10^6 \, Vm^{-1}$.Since the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ 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,$\vec{E} = -1.5 \times 10^6 \hat{i} \, NC^{-1}$.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.

Class 12 Physics Moving Charges and Magnetism Mix Examples-Moving Charges and Magnetism

Two charged particles $A$ and $B$,each of charge $+e$ and masses $12 \, amu$ and $13 \, amu$ respectively,follow a circular trajectory in chamber $X$ after passing through a velocity selector as shown in the figure. Both particles enter the velocity selector with a speed of $1.5 \times 10^6 \, ms^{-1}$. $A$ uniform magnetic field of strength $1.0 \, T$ is maintained within the chamber $X$ and in the velocity selector,directed into the plane ($-z$ direction).

A
Electric field across the conducting plate of the velocity selector is $-10^6 \, NC^{-1} \hat{i}$.
B
Electric field across the conducting plate of the velocity selector is $10^6 \, NC^{-1} \hat{i}$.
C
The ratio $r_A / r_B$ of the radii of the circular paths for the two particles is $12/13$.
D
The ratio $r_A / r_B$ of the radii of the circular paths for the two particles is $13/12$.

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Moving Charges and Magnetism

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Mix Examples-Moving Charges and Magnetism

The temperature at which the thermo-electric power of a thermocouple becomes zero is called:

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Six point charges,each of magnitude $q$,are arranged in different manners as shown in the image. In each case,a point $M$ and a line $PQ$ passing through $M$ are shown. Let $E$ be the electric field and $V$ be the electric potential at $M$ (potential at infinity is zero) due to the given charge distribution when it is at rest. Now,the whole system is set into rotation with a constant angular velocity about the line $PQ$. Let $B$ be the magnetic field at $M$ and $\mu$ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. Match the conditions in Column $I$ with the configurations in Column $II$.
Column $I$ Column $II$
$(A)$ $E=0$ $(p)$ Charges at corners of a regular hexagon. $M$ is the centre. $PQ$ is perpendicular to the plane.
$(B)$ $V \neq 0$ $(q)$ Charges on a line perpendicular to $PQ$ at equal intervals. $M$ is the mid-point.
$(C)$ $B=0$ $(r)$ Charges on two coplanar concentric rings. $M$ is the common centre. $PQ$ is perpendicular to the plane.
$(D)$ $\mu \neq 0$ $(s)$ Charges at corners and mid-points of a rectangle. $M$ is the centre. $PQ$ is parallel to the longer sides.
$(t)$ Charges on two coplanar,identical rings. $M$ is the mid-point between centres. $PQ$ is perpendicular to the line joining centres.

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The energies required to set up in a cube of side $10 \,cm$ $(i)$ a uniform electric field of $10^7 \,Vm^{-1}$ and (ii) a uniform magnetic field of $0.25 \,Wbm^{-2}$ are respectively about $(\mu_0=4 \pi \times 10^{-7} \,Hm^{-1}, \varepsilon_0=8.9 \times 10^{-12} \,Fm^{-1})$

MediumAP EAMCET 2018

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Match the following:
List-$I$ List-$II$
$a$. Fleming's left-hand rule $e$. Direction of induced current
$b$. Fleming's right-hand rule $f$. South pole
$c$. Clockwise current $g$. North pole
$d$. Anticlockwise current $h$. Direction of force

The correct answer is:

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The unit of magnetic flux density (or magnetic induction) is:

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Column $I$	Column $II$
$(A)$ $E=0$	$(p)$ Charges at corners of a regular hexagon. $M$ is the centre. $PQ$ is perpendicular to the plane.
$(B)$ $V \neq 0$	$(q)$ Charges on a line perpendicular to $PQ$ at equal intervals. $M$ is the mid-point.
$(C)$ $B=0$	$(r)$ Charges on two coplanar concentric rings. $M$ is the common centre. $PQ$ is perpendicular to the plane.
$(D)$ $\mu \neq 0$	$(s)$ Charges at corners and mid-points of a rectangle. $M$ is the centre. $PQ$ is parallel to the longer sides.
	$(t)$ Charges on two coplanar,identical rings. $M$ is the mid-point between centres. $PQ$ is perpendicular to the line joining centres.

List-$I$	List-$II$
$a$. Fleming's left-hand rule	$e$. Direction of induced current
$b$. Fleming's right-hand rule	$f$. South pole
$c$. Clockwise current	$g$. North pole
$d$. Anticlockwise current	$h$. Direction of force

Similar Questions

The temperature at which the thermo-electric power of a thermocouple becomes zero is called:

The energies required to set up in a cube of side $10 \,cm$ $(i)$ a uniform electric field of $10^7 \,Vm^{-1}$ and (ii) a uniform magnetic field of $0.25 \,Wbm^{-2}$ are respectively about $(\mu_0=4 \pi \times 10^{-7} \,Hm^{-1}, \varepsilon_0=8.9 \times 10^{-12} \,Fm^{-1})$

Match the following:List-$I$List-$II$$a$. Fleming's left-hand rule$e$. Direction of induced current$b$. Fleming's right-hand rule$f$. South pole$c$. Clockwise current$g$. North pole$d$. Anticlockwise current$h$. Direction of forceThe correct answer is:

The unit of magnetic flux density (or magnetic induction) is:

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Match the following:
List-$I$ List-$II$
$a$. Fleming's left-hand rule $e$. Direction of induced current
$b$. Fleming's right-hand rule $f$. South pole
$c$. Clockwise current $g$. North pole
$d$. Anticlockwise current $h$. Direction of force

The correct answer is: