(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

Medium

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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$A$ uniform magnetic field $B$ exists in a region. An electron of charge $q$ and mass $m$ moving with velocity $v$ enters the region in a direction perpendicular to the magnetic field. Considering Bohr angular momentum quantization, which of the following statement$(s)$ is/are true?

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An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

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$A$ charged particle in a uniform magnetic field $B = B_0 \hat{k}$ starts moving from the origin with velocity $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$. The trajectory of the particle and the time $t$ at which it reaches $2 \text{ m}$ above the $x-y$ plane are,

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$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

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In Thomson's method for determining the $e/m$ ratio of an electron,which of the following conditions is true regarding the fields?

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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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$A$ uniform magnetic field $B$ exists in a region. An electron of charge $q$ and mass $m$ moving with velocity $v$ enters the region in a direction perpendicular to the magnetic field. Considering Bohr angular momentum quantization, which of the following statement$(s)$ is/are true?

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

$A$ charged particle in a uniform magnetic field $B = B_0 \hat{k}$ starts moving from the origin with velocity $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$. The trajectory of the particle and the time $t$ at which it reaches $2 \text{ m}$ above the $x-y$ plane are,

$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

In Thomson's method for determining the $e/m$ ratio of an electron,which of the following conditions is true regarding the fields?

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