3 Marks Question

Question 13 Marks

Write relation between $B , H , I , \chi$ and $\mu$.

Answer

Magnetic suspectibility : In a small magnetised field, intensity of magnetisation (I) of the substance is proportional to magnetic field $( H )$,
which means $I \propto H$ or, $ \begin{array}{c} I=\chi H \\or \chi=\frac{I}{H} \end{array} $
Here, is a constant which is known as magnetic susceptibility of the substance.
Magnetic permeability : Capacity of passing of magnetic field lines from any medium is known as magnetic permeability of that medium. Mathematically, it is the ratio of magnetic induction induced B inside the substance and magnetic field H . It is represented by $\mu$. It is a scalar quantity.
$ \mu=\frac{B}{H} $ On dividing equation (1) by (2), $ \frac{\mu}{\chi}=\frac{\frac{B}{H}}{\frac{I}{H}} $

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Question 23 Marks

What is magnetic flux density? Obtain a relation between magnetic flux and magnetic induction.

Answer

Number of magnetic field lines passing through surface of unit area placed perpendicular to the force lines is called magnetic flux density. It is a vector quantity which is denoted by $\overrightarrow{ B }$. It is also called magnetisation field or intensity of magnetisation field or magnetic induction. Its unit is weber/ meter $^2$ or tesla or newton/ampere-metre.
If in the magnetic field of intensity $\vec{B}$, a surface of infinitely small surface is located at any point, then magnetic flux associated with it will be $\delta \phi_{ B ^{\prime}}$
$ \delta \phi_{B}=\overrightarrow{B} \cdot \overrightarrow{\delta S}=B \delta s \cos \theta $
And total associated magnetic flux $\phi_{ B }=\int_{ S } \overrightarrow{ B } . \overrightarrow{\delta S }$
Magnetic flux is a scalar quantity but magnetic flux density is a vector quantity.

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Question 33 Marks

Define a magnet and write its characteristics of directive property and pole strength.

Answer

Such a substance which attracts other magnetic substances such as iron etc. towards itself and when it is suspended freely, then it rests in north-south direction. Its two and equal opposite poles are separated by some distance.
Directive property : When any bar magnet is suspended freely, it always rests in a definite direction. Its one end is towards the north and the other is towards the south which are known as north pole and south pole respectively. A bar magnet always exhibits north and south direction.
Pole strength : Strength of magnetisation of magnetic poles is known as pole strength. Pole strength of both poles of a bar magnet is same. If a bar magnet is cut perpendicular to its axis, then pole strength of poles of each part will almost be same as of original magnet and if it is cut in to equal parts along its axis, then pole strength of each part will be half that of original bar magnet.

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Question 43 Marks

Magnetic moment of a small bar magnet is 0.24 newton metre tesla ${ }^{-1}$. Calculate magnetising field of the magnet at a distance of 10 cm from the centre of the magnet on its axis.

Answer

Given : $m=0.24$ newton metre tesla ${ }^{-1}$
$ r=10 cm=0.10 \text { metre } $
$\therefore$ In axial position, $ B=\frac{\mu_0}{4 \pi}\left(\frac{2 m}{r^3}\right) $
On putting the values, $B =\frac{4 \pi \times 10^{-7}}{4 \pi}\left(\frac{2 m}{r^3}\right)$
$=10^{-7}\left(\frac{2 m}{r^3}\right)=\frac{10^{-7}(2 \times 0.24)}{(0.10)^3}$
$=4.8 \times 10^{-5}$ tesla $\quad$ Ans.

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Question 53 Marks

0.5 A of current is flowing through a coil of 25 turns and area $200 cm^2$. What is the pole strength of 10 cm length of bar magnet whose magnetic moment is equal to magnetic moment of the coil?

Answer

Magnetic moment of bar magnet $=$ Magnetic moment of the coil
$ \therefore \quad q_m \times 2 l=\text { NIA } $
Here, pole strength $ \left(q_m\right)=\frac{\text { NIA }}{2 l} $
On putting the values,
$ \begin{aligned} q_m & =\frac{25 \times 0.5 \times\left(200 \times 10^{-4}\right)}{10 \times 10^{-2}} A-m \\ & =2.5 A-m . \end{aligned} $

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Question 63 Marks

A coil of area $5 cm^2$ is kept in a uniform magnetising field of $2.5 N / Am$. Number of turns in the coil is 100 . If 0.2 A of current is flowing, then calculate:
(a) Magnetising dipole moment,
(b) Maximum moment of force.

Answer

Given : Area of the bottom of the coil $A =5 cm^2=$ $5 \times 10^{-4} m^2$, Magnetic field intensity B $=2.5 N / Am$, Number of turns, $N =100$, Current flowing in the coil $I =0.2 A$
(a) Magnetising dipole moment $m=$ NIA $ \begin{aligned} m & =100 \times 0.2 A \times 5 \times 10^{-4} m^2 \\ & =0.01 A-m^2 \end{aligned} $
(b) Magnetic moment of force $ \begin{aligned} \tau_{\max } & =m B \\ & =0.01 A-m^2 \times 2.5 N / A m \\ & =0.025 N-m \end{aligned} $

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