Q175 / 200 · AP EAMCET 2001 · Physics

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

  • A
    zero
  • B
    $10 \mathrm{~T}$
  • Correct answer
    C
    $20 \mathrm{~T}$
  • D
    $200 \mathrm{~T}$

Solution

$\begin{gathered}m=0.6 \mathrm{~g}=0.6 \times 10^{-3} \mathrm{~kg}, q=25 \mathrm{nC}=25 \times 10^{-9} \mathrm{C} \\ v=1.2 \times 10^4 \mathrm{~m} / \mathrm{s}\end{gathered}$ The particle is moving with uniform velocity (acceleration is zero), so magnetic force will be balancing the weight of the particle. $m g=B q v$ $\begin{aligned} 0.6 \times 10^{-3} \times 10 & =B \times 25 \times 10^{-9} \times 1.2 \times 10^4 \\ B & =\frac{0.6 \times 10^{-3} \times 10}{25 \times 10^{-9} \times 1.2 \times 10^4} \\ & =\frac{6 \times 10^{-3}}{30 \times 10^{-5}}=\frac{6 \times 10^2}{30} \\ & =20 \mathrm{~T}\end{aligned}$

Hence, the correct answer is option: 'C'.

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