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A particle of charge $-16 \times 10^{-18}$ coulomb moving with velocity $10 \mathrm{~m} \mathrm{~s}^{-1}$ along the $X$-axis enters a region, where a magnetic field of induction $B$ is along the $Y$-axis and an electric field of magnitude $10^{4} \mathrm{~V}^{-1}$ is along the negative $Z$-axis. If the charged particle continues moving along the $X$-axis, the magnitude of $B$ is

(a) $10^{3} \mathrm{~Wb} \mathrm{~m}^{-2}$

(b) $10^{5} \mathrm{~Wb} \mathrm{~m}$

(c) $10^{16} \mathrm{~Wb} \mathrm{~m}^{-2}$

(d) $10^{-2} \mathrm{~Wb} \mathrm{~m}^{-2}$

The correct option of this question will be (a).

Solution —

Given $q=-16 \times 10^{-18} C , v=10 \hat{ m s } s ^{-1}$

$\vec{B} =B \hat{j}, \vec{E}=-10^{4} \hat{k} Vm ^{-1}$

$\vec{F}_{e} =q \vec{E}=-16 \times 10^{-18} \times\left(-10^{4} \hat{k}\right)$

=$16 \times 10^{-14} \hat{k} N$

$\vec{F}_{m} =q(\vec{v} \times \vec{B})=-16 \times 10^{-18}(10 \hat{i} \times B \hat{j})$

=$-16 \times 10^{-17} B \hat{k} N$

As the particle continues to move along the same direction

$F_{m}=F_{e}$

$\therefore \quad 16 \times 10^{-17}$

$B$ = $16 \times 10^{-14}$

$\text { or } B=10^{3} Wb m ^{-2}$

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