SATHEE: moving-charges-and-magnetism Question 11

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,

$$ \begin{align*} \vec{F} & =\vec{F}{E}+\overrightarrow{F{m}}=\overrightarrow{0} \ \overrightarrow{F_{E}} & =-\overrightarrow{F_{m}} \tag{1} \end{align*} $$

$\overset{―}{F_{E}} \to$ electric force, $\vec{F_{m}} \to$ 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)

$\begin{matrix} (5) & E = \frac{μ_{0}}{2 π} \frac{v I}{r} \end{matrix}$

Magnetic force $F_{m}$ is towards wire.

$∴$ 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 $\vec{d B}$ , due to a current element $\vec{d l}$ , 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 $\vec{d l}$ and $\vec{r}$ .

$1 / 2$

In vector notation,

$\vec{d B} = \frac{μ_{0}}{4 π} I \frac{\vec{d l} \times \vec{r}}{r^{3}}$

(b)

We have, $\begin{aligned} \vec{d B} & = \frac{μ_{0}}{4 π} I \frac{| \vec{d I} \times \vec{r} |}{r^{3}} r^{2} & = x^{2} + R^{2} ∴ d B & = \frac{μ_{0} I}{4 π} \frac{d l}{{(x^{2} + R^{2})}^{3 / 2}} \end{aligned}$

We need to add only the components of $d \bar{B}$ along the axis of the coil.

Hence,

$\begin{aligned} B & = \int \frac{μ_{0} I}{4 π} \cdot \frac{I d l}{{(x^{2} + R^{2})}^{3 / 2}} \cos θ & = \int \frac{μ_{0} I}{4 π} \frac{(I d l) R}{{(x^{2} + R^{2})}^{3 / 2}} & = \frac{μ_{0} I R^{2}}{2 {(x^{2} + R^{2})}^{3 / 2}} \vec{B} & = \frac{μ_{0} I R^{2}}{2 {(x^{2} + R^{2})}^{3 / 2}} \hat{i} \end{aligned}$

[CBSE Marking Scheme 2018]

Moving Charges And Magnetism

moving-charges-and-magnetism Question 11

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Moving Charges And Magnetism

moving-charges-and-magnetism Question 11

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu