Question

Figure shows a snapshot of a proton moving at velocity$\overrightarrow{\mathbf{v}}\mathbf{}\mathbf{=}\left(\mathbf{200}\mathbf{}\frac{\mathbf{m}}{\mathbf{s}}\right)\hat{\mathbf{j}}$toward a long straight wire with current $\mathbf{i}\mathbf{=}\mathbf{350}\mathbf{}\mathbf{mA}$. At the instant shown, the proton’s distance from the wire is $\mathbf{d}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{89}\mathbf{}\mathbf{cm}$. In unit-vector notation, what is the magnetic force on the proton due to the current?

Short answer

The magnitude of magnetic force on proton is $\left(-7.75\times {10}^{-23}\text{N}\right)\hat{i}$.

Step-by-step solution

Given

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

Determine the concept and the formulas:

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

$B=\frac{{\mu }_{0}i}{2\pi r}$

Magnetic force, $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$.

Calculate the magnitude of magnetic force on proton

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

$B=\frac{{\mu }_{0}i}{2\pi r}$

$B=\frac{4\pi \times {10}^{-7}\times .350}{2\times 3.14\times 2.89\times {10}^{-2}}$

$B=2.42\times {10}^{-6}T$

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

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

Now, force is

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

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

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