SATHEE: moving-charges-and-magnetism Question 14

moving-charges-and-magnetism Question 14

Question: Q. 6. Using Biot-Savart’s law, deduce the expression for the magnetic field at a point $(x)$ on the axis of a circular current carrying loop of radius R. How is the direction of the magnetic field determined at this point?

U] [Foreign I, II 2017]

Show Answer

Solution:

Ans. Try yourself, Similar to Q. 3 (b) of Short Answer.

[AI Q. 7. (i) Write the expression for the magnetic force acting on a charged particle moving with velocity $v$ in the presence of magnetic field $B$ .

(ii) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer.

U [Dehhi, II, III 2016]

Ans.

(i)

(Give Full credit of this part even if a student writes :

$F = q v B \sin θ$

and force $(\vec{F})$ acts perpendicular to the plane containing $\vec{v}$ and $\vec{B}$ )

(ii)

$1 / 2 + 1 / 2 + 1 / 2$

$α$

$\overset{x}{\to} x x x$

Justification : Direction of force experienced by the particle will be according to the Fleming’s left hand rule/(any other alternative correct rule.) 1/2

[CBSE Marking Scheme 2016]

Detailed Answer :

(i) A charged particle with charge $q$ , moving with velocity $\vec{v}$ , in a magnetic field having field strength $\vec{B}$ , then the forceacting on it is given as :

$\begin{aligned} \vec{F} = q (\vec{v} \times \vec{B}) & \vec{F} = \vec{F} q v B \sin θ \hat{n} \end{aligned}$

where, $θ =$ angle between velocity and magnetic field

Here direction of force is shown by cross product of

(ii) velocity and ${\underset{x}{x}}^{x_{x}}_{x}$ .

Here, $α$ -particle will trace the circular path in a clockwise direction, since its deviation is in the direction of $\vec{v} \times \vec{B}$ .

$1 / 2$

Neutron will pass without any deviation, as magnetic field, exerts no force on neutral particle. $1 / 2$ Electron will trace circular path in an anticlockwise direction, since its deviation is opposite to $\vec{v} \times \vec{B}$ with radius $r = \frac{m v}{q B}$ .

$1 / 2$

It is further observed that the direction of force experienced by the particle will be as per Fleming’s left hand rule.

Moving Charges And Magnetism

moving-charges-and-magnetism Question 14

Detailed Answer :

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Moving Charges And Magnetism

moving-charges-and-magnetism Question 14

Detailed Answer :

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

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