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