An electric ﬁeld of $1500 V m^{- 1}$ and a magnetic ﬁeld of $0.4 Wb / m^{2}$ act on a moving electron. The minimum uniform speed along a straight line the electron could have so that it doesnot deviate from its path is

$1.6 \times 10^{15} m / s$

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$6 \times 10^{- 16} m / s$

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$3.75 \times 10^{3} m / s$

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$3.75 \times 10^{2} m / s$

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Solution

The correct option is C
$3.75 \times 10^{3} m / s$

For the electron to move undeviated
$F_{n e t} = F_{m} - F_{e} = 0$ ,

$F_{m} = F_{e}$
Where $F_{m}$ is force due to magnetic field and
$F_{e}$ is force due to electric field

For minumum speed of the electron,
component of $\to B$ (Magnetic Field) perpendicular to $\to v$ (Velocity of electron)should be maximum

So angle between $\to B$ and $\to v$ should be $90^{\circ}$
$\Rightarrow e v B S i n (90^{\circ}) = e E$

or $v = \frac{E}{B} = \frac{1500}{0.40} = 3.75 \times 10^{3} m / s$

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An electric ﬁeld of 1500 Vm−1 and a magnetic ﬁeld of 0.4 Wb/m2 act on a moving electron. The minimum uniform speed along a straight line the electron could have so that it doesnot deviate from its path is

An electric ﬁeld of $1500 V m^{- 1}$ and a magnetic ﬁeld of $0.4 Wb / m^{2}$ act on a moving electron. The minimum uniform speed along a straight line the electron could have so that it doesnot deviate from its path is