$A$ velocity selector is to be constructed to select ions with a velocity of $6 \,km \,s^{-1}$. If the electric field used is $400 \,V \,m^{-1}$, then the magnetic field to be used is

  • A
    $\frac{11}{20} \,T$
  • B
    $\frac{2}{3} \,T$
  • C
    $\frac{1}{15} \,T$
  • D
    $\frac{2}{15} \,T$

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Similar Questions

$A$ particle of mass $0.6\, g$ and having a charge of $25\, nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4\, m/s$ in a uniform magnetic field. If the particle continues to move with uniform velocity,the value of the magnetic induction is $(g = 10\, m/s^2)$.

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. The value of the magnetic induction is $\left(g=10 \,ms^{-2}\right)$.

$H^+$,$He^+$,and $O^{++}$ are projected in a uniform transverse magnetic field with equal accelerating potential. If their masses are $1 \, a.m.u.$,$4 \, a.m.u.$,and $16 \, a.m.u.$ respectively,find the ratio of their radii.

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$A$ proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha$-particle (mass = $4m$ and charge = $+2e$) so that it can revolve in the path of the same radius?

Assertion $(A)$: When a proton and a neutron enter into a transverse magnetic field with equal speeds,they trace circular paths of equal radii.
Reason $(R)$: In a transverse magnetic field,the period of revolution of a charged particle in a circular path is directly proportional to the mass of the particle.

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