Question

Q:1 A long straight wire XY carries a current 3.0A. A proton P
moves at 3×10^-6 ms^-1
parallel to the wire 0.15m away from it.Find the magnitude and direction of the
force exerted by the magnetic field on the proton.​

Answer 1

ANSWER:-

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• Given, that the Proton P moves parallel to the wire.

•The attachment will help us to understand the Question better.

Magnetic field due to the wire at P,

$\bf{\large{ B={\frac{{u}_0 2i}{4 \pi r}}}}$$\: \: \: \: \: \:{[For \: a \: long \: conductor]}$

But,

$i=3.0A$

$r=0.15m$

$B= {\frac{2 \times {10^{-7}} \times 3}{0.15}}$ $\: \: \: \: \: {\large{[\frac{{u}_0}{4 \pi }}={10}^{-7}]}$

$\bf{B= 4 \times 10 ^{-6} T}$

Force experienced by the proton in this magnetic field,

$\bf{\implies F = qv B sin \theta}$

Here,

$q = 1.6 \times 10^{-19} C \\ v = 3 \times 10 ^{-6}ms^{-1} \\ \theta=90 \degree$

Therefore,

$F ={ 1.6 \times {10^{-19}} \times 3 \times {10 ^{-6}} \times 4 \times {10 ^{-6}} \times 1 N}$

$=19.2 \times 10^{-31}$

$\bf{=1.92 \times 10^{-30}N}$

• According to Fleming’s left hand rule the direction of force will be towards the wire XY.

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Note:-

• Mate, as the correction in the Question provided by you. We have the Velocity of Proton as: $\bf{v = 3 \times 10 ^{6}ms^{-1}}$.

•$\bf{Again \: Solving\: it:-}$

• We have:-

$F ={ 1.6 \times {10^{-19}} \times 3 \times {10 ^{6}} \times 4 \times {10 ^{-6}} \times 1 N}$

$=19.2 \times 10^{-19}$

$\bf{=1.92 \times 10^{-18}N}$

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Formula Used:-

• $\bf{\large{ B={\frac{{u}_0 2i}{4 \pi r}}}}$

• $\bf{F = qv B sin \theta}$

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Answer 2

hope it will be helpful to you.