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Standard X

Physics

Magnetic Force on Current Carrying Wire

In the Millik...

Question

In the Millikan's oil drop experiment the oil drop is subjected to a horizontal electric field of 2 N/C and the drop moves with a constant velocity making an angle of $45^{\circ}$ with the horizontal. If the weight of the drop is W, then the electric charge, in coulomb, on the drop is

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\frac{W}{2}

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\frac{W}{4}

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\frac{W}{8}

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Solution

The correct option is B $\frac{W}{2}$
From the free body diagram and given resultant force direction, i.e., at $45^{0}$ to horizontal
$F_{e} = F_{g}$ , $F_{e} a n d F_{g}$ are the electric and gravitational forces on drop
So, $e E = m g$ and substituting the given values

$E = 2 \frac{N}{C}$ ,

$m g = W$

We get, $e = \frac{W}{2}$

So, the answer is option (B).

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In the Millikan's oil drop experiment the oil drop is subjected to a horizontal electric field of 2 N/C and the drop moves with a constant velocity making an angle of 45∘ with the horizontal. If the weight of the drop is W, then the electric charge, in coulomb, on the drop is

The correct option is B W2From the free body diagram and given resultant force direction, i.e., at 450 to horizontalFe=Fg , FeandFg are the electric and gravitational forces on dropSo, eE=mg and substituting the given valuesE=2NC , mg=WWe get, e=W2So, the answer is option (B).

In the Millikan's oil drop experiment the oil drop is subjected to a horizontal electric field of 2 N/C and the drop moves with a constant velocity making an angle of $45^{\circ}$ with the horizontal. If the weight of the drop is W, then the electric charge, in coulomb, on the drop is