91Ó°ÊÓ

1. Permeability of free space, ${\mathrm{μ}}_0=4\mathrm{π}×{10}^{-7}\frac{\text{Tm}}{\text{A}}$.
2. Given distance, $d=2.89\text{cm}$.
3. Charge on proton, $q=1.6×{10}^{-19}$.
4. Velocity of particle, $v=\left(-200\hat{j}\right)$.
5. Current carried by wire,$i=350\text{mA}$.

Calculate the magnetic field due to the long wire at the position of proton. Use this value in the equation of magnetic force to calculate it.

Formula:

Magnetic field

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

Magnetic force, $\stackrel{→}{F}=q\stackrel{→}{v}×\stackrel{→}{B}$.

First of all, we have to find the magnetic field B.

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

$B=\frac{4\mathrm{π}×{10}^{-7}×.350}{2×3.14×2.89×{10}^{-2}}$

$B=2.42×{10}^{-6}T$

According to right hand rule, we can say that the magnetic field is out of paper inlocalid="1663227168373" $+k$direction.

$B=\left(2.4221×{10}^{-6}T\right)\hat{k}$

Now, force is

$\stackrel{→}{F}=q\stackrel{→}{v}×\stackrel{→}{B}$

$\stackrel{→}{F}=\left(1.6×{10}^{-19}\right)\left(-200\hat{j}\right)×\left(2.42×{10}^{-6}\hat{k}\right)$

localid="1663227211827" $\stackrel{→}{F}=\left(-7.75×{10}^{-23}N\right)\hat{i}$