Question 266: Magnetism and Matter

Question

A 5 mg particle carrying a charge of $5 \pi 	imes 10^{-6} \mathrm{C}$ is moving with velocity of $(3 \hat{i}+2 \hat{k}) 	imes 10^{-2} \mathrm{~m} / \mathrm{s}$ in a region having magnetic field $\vec{B}=0.1 \hat{k} \mathrm{~Wb} / \mathrm{m}^2$. It moves a distance of $\alpha$ meter along $\hat{k}$ when it completes 5 revolutions. The value of $\alpha$ is $\_\_\_\_$。

Accepted Answer

When a charged particle enters a magnetic field with a velocity that has both a component perpendicular to the field and a component parallel to it, the particle follows a helical path.
The perpendicular component $\left(\mathrm{v}_{\perp}ight)$ causes circular motion due to the Lorentz force $\left(\mathrm{F}=\mathrm{qv}_{\perp} \mathrm{B}ight)$.
The parallel component ( $\mathrm{v}_{\|}$) remains constant because the magnetic force on it is zero ( $\mathrm{F}=\mathrm{qv}_{\|} \mathrm{B} \sin 0^{\circ}=0$ ), causing the particle to move linearly along the field lines.
The velocity is given as $\vec{v}=(3 \hat{\imath}+2 \hat{k}) 	imes 10^{-2} \mathrm{~m} / \mathrm{s}$ and the magnetic field is $\vec{B}=0.1 \hat{\mathrm{k}} \mathrm{T}$.
The component along the z -axis $(\widehat{\mathrm{k}})$ is $2 	imes 10^{-2} \mathrm{~m} / \mathrm{s}$.
The component along the x -axis (ÃÂ®) is $3 	imes 10^{-2} \mathrm{~m} / \mathrm{s}$.
For one revolution, the time taken is determined by the circular component of the motion.
Centripetal force is provided by the magnetic force:
$$ \frac{m v_{\perp}^2}{r}=q v_{\perp} B \Rightarrow r=\frac{m v_{\perp}}{q B} $$
The time for one revolution is.
$$ \mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}_{\perp}}=\frac{2 \pi \mathrm{~m}}{\mathrm{qB}} $$
The distance moved along the field lines in 1 revolution is called the Pitch.
$$ 	ext { Pitch }=\mathrm{v}_{\|} 	imes \mathrm{T} $$
For 5 revolutions, the total distance (a) is:
$$ \mathrm{a}=5 	imes \mathrm{v}_{\|} 	imes \mathrm{T}=5 	imes \mathrm{v}_{\|} 	imes\left(\frac{2 \pi \mathrm{~m}}{\mathrm{qB}}ight) $$
The mass of charged particle is $\mathrm{m}=5 \mathrm{mg}=5 	imes 10^{-6} \mathrm{~kg}$.
The magnitude of charge on the particle is $\mathrm{q}=5 \pi 	imes 10^{-6} \mathrm{C}$
The parallel component of velocity to magnetic field is $\mathrm{v}_{\|}=2 	imes 10^{-2} \mathrm{~m} / \mathrm{s}$
The magnitude of magnetic field in the region is $\mathrm{B}=0.1 \mathrm{~T}$
So, the distance travelled is;
$$ a=5 	imes\left(2 	imes 10^{-2}ight) 	imes \frac{2 \pi 	imes\left(5 	imes 10^{-6}ight)}{\left(5 \pi 	imes 10^{-6}ight) 	imes 0.1} $$
$$\Rightarrow $$ $\mathrm{a}=10 	imes 10^{-2} 	imes \frac{10}{5 	imes 0.1}$
$$\Rightarrow $$ $\mathrm{a}=10^{-1} 	imes \frac{10}{0.5}$
$$\Rightarrow $$ $$ a=0.1 	imes 20=2 	ext { meters } $$
Therefore, the distance a moved along the $\widehat{k}$ direction is 2 meters.

Magnetism and Matter

Integer / Numeric Answer Type