# A long straight wire AB carries a current of 3 A. A proton P travels at  $5\times 10^{6}m/s$ parallel to the wire 0·3 m from it and in a direction opposite to the current as shown in the figure. Calculate the force which the magnetic field due to the current carrying wire exerts on the proton. Also give its direction.

Given that,

$r=0.3m$

Current  $I=3A$

Speed of proton $= 5\times 10^{6}m/s$

The magnetic field produced by long wire AB at the point P is

$B=\frac{\mu_0I}{2\pi r}$

Proton is moving in perpendicular to the magnetic field so the magnetic force is

$F_m=qvB$

Then,

$B=\frac{\mu _0I}{2\pi r}=\frac{4\pi \times 10^7\times 3}{2\times \pi \times 0. 3}=2\times 10^{-6}T$

So,

$F_m=qvB$

$=1.6\times 10^{-19}\times 5\times 10^6\times 2\times 10^{-6}$

$=1.6\times 10^{-18}N$

By fleming's left-hand rule, the direction of force is away from the long wire AB towards the right.

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