(B) The radius of a circular path for a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.Since the particle is accelerated through

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$\left(\frac{R_1}{R_2}\right)^2$

(B) The radius of a circular path for a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.Since the particle is accelerated through a potential difference $V$,its kinetic energy is $KE = qV = \frac{1}{2}mv^2$,which implies $v = \sqrt{\frac{2qV}{m}}$.Substituting $v$ into the radius formula: $R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Since $q$,$V$,and $B$ are constant for both particles,we have $R \propto \sqrt{m}$,which implies $m \propto R^2$.Therefore,the ratio of the masses is $\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2$.

Class 12 Physics Moving Charges and Magnetism Motion of Charged Particle In Magnetic Field

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Two particles $X$ and $Y$ having equal charges are being accelerated through the same potential difference. Thereafter,they enter normally into a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $X$ and $Y$ is:

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A
$\left(\frac{R_2}{R_1}\right)^2$
B
$\left(\frac{R_1}{R_2}\right)^2$
C
$\frac{R_1}{R_2}$
D
$\frac{R_2}{R_1}$

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

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An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved,then the resulting path would have a radius of:

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$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

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When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be:
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix

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$A$ proton accelerated by a potential difference of $500 \ kV$ moves through a transverse magnetic field of $0.5 \ T$ as shown in the figure. The angle $\theta$,through which the proton deviates from the initial direction of its motion,is (mass of proton $= 1.6 \times 10^{-27} \ kg$):- (in $^{\circ}$)

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$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

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Two particles $X$ and $Y$ having equal charges are being accelerated through the same potential difference. Thereafter,they enter normally into a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $X$ and $Y$ is:

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An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved,then the resulting path would have a radius of:

$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be:$(1)$ a straight line$(2)$ a circle$(3)$ a helix

$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

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When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be:
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix