SATHEE: magnetism-and-matter Question 21

magnetism-and-matter Question 21

Question: Q. 6. (a) An iron ring of relative permeability $μ_{r}$ has windings of insulated copper wire of $n$ turns per metre. When the current in the windings is $I$ , find the expression for the magnetic field in the ring.

(b) Thesusceptibility of a magnetic material is 0.9853 . Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.

R [Delhi & OD, 2018]

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Solution:

Ans. (a) Expression for Ampere’s circuital law $1 / 2$

Derivation of magnetic field inside the ring

$1 / 2$

(b) Identification of the material

$1 / 2$

Drawing the modification of the field pattern $1 / 2$

(a) From Ampere’s circuital law, we have,

$\begin{matrix} (i) & \oint \vec{B} \cdot d \vec{l} = μ_{0} μ_{r} I_{enclosed} \end{matrix}$

For the field inside the ring, we can write

$\begin{aligned} \oint \vec{B} \cdot d \vec{l} = \oint B d l = B \cdot 2 π r & (r = radius of the ring) & Also, I_{enclosed} = (2 π r n) I \end{aligned}$

using equation (i)

$∴ B \cdot 2 π r = μ_{0} μ_{r} (2 π r n) I$

$∴ B = μ_{0} μ_{r} n I$

[Award these $(\frac{1}{2} + \frac{1}{2})$ marks even if the result is

written without giving the derivation] $1 / 2$

(b) The material is paramagnetic. $1 / 2$

The field pattern gets modified as shown in the figure below.

$1 / 2$

[CBSE Marking Scheme 2018]

Detailed Answer :

(a)

Apply Ampere’s Law for the magnetic field due to iron ring wounded by insulating copper wire, having current I, $\oint \bar{B} \cdot \overset{―}{d l} = μ^{'} \times ($ current enclosed by closed path $) 1 / 2$

or, $B d l \cos 0^{\circ} = μ^{'} \times (n \times 2 \times r) \times I$

or, $B \times 2 \times r \times 1 = μ^{'} n \times 2 \times r \times I$

or, $B = μ^{'} n I$

But $μ_{r} = \frac{μ^{'}}{μ_{o}}$

So, $B = μ_{0} μ_{r} n I$

This is the required expression for magnetic field. Where, $μ_{r} \to$ relative permeability, $μ_{0} \to$ permeability of free space, $n \to$ number of turns per unit length $n$

(b) Given : susceptibility $χ = 0.9853$ since susceptibility $χ$ given is + ve and less than unity i.e. $χ < + 1 1 / 2$ $\Rightarrow$ magnetic material is paramagnetic material. Thus when paramagnetic material is placed in the uniform magnetic field then the modified magnetic field is shown in figure.

Paramagnetic material

Magnetism And Matter

magnetism-and-matter Question 21

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Magnetism And Matter

magnetism-and-matter Question 21

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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