moving-charges-and-magnetism Question 11
Question: Q. 2. A particle of mass $m$ and charge $q$ is in motion at speed $v$ parallel to a long straight conductor carrying current $I$ as shown below.
Find magnitude and direction of electric field required so that the particle goes undeflected.
R&U [CBSE SQP 2018]
For the charged particle to move undeflected
Net force,
$$ \begin{align*} \vec{F} & =\vec{F}{E}+\overrightarrow{F{m}}=\overrightarrow{0} \ \overrightarrow{F_{E}} & =-\overrightarrow{F_{m}} \tag{1} \end{align*} $$
$\overline{F_{E}} \rightarrow$ electric force, $\overrightarrow{F_{m}} \rightarrow$ magnetic force
$$ \begin{align*} \left|\vec{F}{E}\right| & =\left|\vec{F}{m}\right| \tag{2}\ q E & =B q v \sin 90^{\circ}=B q v \ E & =v B \ B & =\frac{\mu_{0} I}{2 \pi r} \tag{4} \end{align*} $$
Using (4) and (3)
$$ \begin{equation*} E=\frac{\mu_{0}}{2 \pi} \frac{v I}{r} \tag{5} \end{equation*} $$
Magnetic force $F_{m}$ is towards wire.
$\therefore$ Electric force and electric field should be away from the line.
[CBSE Marking Scheme 2018]
AI Q. 3. (a) State Biot-Savart’s law and express it in the vector form.
(b) Using Biot-Savart’s law, obtain the expression for the magnetic field due to a circular coil of radius $r$, carrying a current $I$ at a point on its axis distant $x$ from the centre of the coil.
R&U [CBSE Comptt. I, II, III 2018]
$\begin{array}{cc}\text {
Show Answer
Solution:
Ans. (a) Statement of Biot-Savart law } & 1 / 2 \ \text { Its vector form } & 1 / 2\end{array}$
(b) Obtaining the required expression
$1 / 2$ 2
(a) According to Biot Savart law :
The magnitude of magnetic field $\overrightarrow{d B}$, due to a current element $\overrightarrow{d l}$, is
(i) proportional to current $I$ and element length, $d l$
(ii) inversely proportional to the square of the distance $r$.
Its direction is perpendicular to the plane containing $\overrightarrow{d l}$ and $\vec{r}$.
$1 / 2$
In vector notation,
$$ \overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} I \frac{\overrightarrow{d l} \times \vec{r}}{r^{3}} $$
(b)
We have, $\quad \begin{aligned} \overrightarrow{d B} & =\frac{\mu_{0}}{4 \pi} I \frac{|\overrightarrow{d I} \times \vec{r}|}{r^{3}} \ r^{2} & =x^{2}+R^{2} \ \therefore \quad d B & =\frac{\mu_{0} I}{4 \pi} \frac{d l}{\left(x^{2}+R^{2}\right)^{3 / 2}}\end{aligned}$
We need to add only the components of $d \bar{B}$ along the axis of the coil.
Hence,
$$ \begin{aligned} B & =\int \frac{\mu_{0} I}{4 \pi} \cdot \frac{I d l}{\left(x^{2}+R^{2}\right)^{3 / 2}} \cos \theta \ & =\int \frac{\mu_{0} I}{4 \pi} \frac{(I d l) R}{\left(x^{2}+R^{2}\right)^{3 / 2}} \ & =\frac{\mu_{0} I R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} \ \vec{B} & =\frac{\mu_{0} I R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} \hat{i} \end{aligned} $$
[CBSE Marking Scheme 2018]
