(B) The radius $R$ of a charged particle moving in a uniform magnetic field $B$ with velocity $v$ is given by $R = \frac{mv}{qB}$.Since $v$ and $B$ ar

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(B) The radius $R$ of a charged particle moving in a uniform magnetic field $B$ with velocity $v$ is given by $R = \frac{mv}{qB}$.Since $v$ and $B$ are constant for all particles, $R \propto \frac{m}{q}$.For a proton $(p)$: $m_p = m$, $q_p = e$, so $R_p \propto \frac{m}{e}$.For an electron $(e)$: $m_e \approx \frac{m}{1836}$, $q_e = e$, so $R_e \propto \frac{m}{1836e}$.For an $\alpha$-particle: $m_{\alpha} \approx 4m$, $q_{\alpha} = 2e$, so $R_{\alpha} \propto \frac{4m}{2e} = 2\frac{m}{e}$.Comparing the ratios: $R_{\alpha} = 2(\frac{m}{e})$, $R_p = 1(\frac{m}{e})$, $R_e \approx 0.0005(\frac{m}{e})$.Thus, $R_{\alpha} > R_p > R_e$.

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

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$A$ proton, an electron, and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$, and $R_{\alpha}$ are the radii of the circular paths of these particles, then:

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A
$R_{\alpha} = R_p = R_e$
B
$R_{\alpha} > R_p > R_e$
C
$R_{\alpha} < R_p < R_e$
D
$R_{\alpha} > R_p = R_e$

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

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$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes $3$ times the original, the new radius will be

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$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

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$A$ proton and an alpha particle moving with energies in the ratio $1: 4$ enter a uniform magnetic field of $3 \ T$ at right angles to the direction of the magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is

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Given the fact that:
$A)$ Magnetic field exerts force only on a moving charge.
$B)$ Electric field exerts force on both stationary and moving charge.
$C)$ Magnetic field exerts force on a charge moving parallel to the direction of the field.
Which of the following is true?

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$A$ proton of mass $m$ moving with a speed $v$ ($v << c$, where $c$ is the velocity of light in vacuum) completes a circular orbit in time $T$ in a uniform magnetic field. If the speed of the proton is increased to $\sqrt{2}v$, what will be the time needed to complete the circular orbit?

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$A$ proton, an electron, and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$, and $R_{\alpha}$ are the radii of the circular paths of these particles, then:

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$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes $3$ times the original, the new radius will be

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

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Given the fact that:
$A)$ Magnetic field exerts force only on a moving charge.
$B)$ Electric field exerts force on both stationary and moving charge.
$C)$ Magnetic field exerts force on a charge moving parallel to the direction of the field.
Which of the following is true?