Case study (4 Marks)

Question 14 Marks

Mention the changes occurring in the geomagnetic components.

Answer

Changes in the geomagnetic components : Magnitudes of geomagnetic components are not only different at different places on the earth and there are usual and unusual changes in their magnitudes at same place also. These changes are as follows:
(1) Long term changes: According to Kelvin, changes in magnitudes of geomagnetic components occur very slowly with time and after one thousand years, they again come back to their original value. According to him, the cause of this change is the spinning of magnetic poles around the geographic poles.
(2) Annual changes: These changes complete their cycle in one year. These changes are maximum in a particular month and minimum in another particular month and their characteristics is of opposite types in both hemispheres. For example, in the month of February, angle of dip is maximum in northern hemisphere and minimum in southern hemisphere. In contract to this, in the month of August, it is minimum in northern hemisphere and maximum in southern hemisphere.
(3) Daily changes : In magnitude of geomagnetic elements, there are daily changes. These changes are maximum at a particular time of the day and minimum at another particular time of the day. This time is different for three components.
(4) Non-periodic changes : Sometimes there are sudden changes in geomagnetic components which are known as magnetic disturbance. Whenever black spot in the sun is towards earth, then there is sudden change in geomagnetic elements.

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Question 24 Marks

A bar magnet whose area of cross-section is $0.25 cm^2$ is kept in a magnetising field of intensity of $5000 A / m$. If magnitude of magnetic flux passing through the bar is $2.5 \times 10^{-5}$ weber, then calculate :(a) Magnetic induction (b) Magnetic susceptibility (c) Intensity of magnetisation.

Answer

(a) Area of cross-section
$ A=0.25 cm^2=0.25 \times 10^{-4} m^2=25 \times 10^{-6} m^2 $
Magnetic flux passing through cross-section of the bar, $ \phi_{B}=2.5 \times 10^{-5} $
$\therefore$ Magnetic induction,
$ B=\frac{\phi_{B}}{A}=\frac{2.5 \times 10^{-5}}{2.5 \times 10^{-6}}=1 weber / m^2 $ Ans.
(b) Magnetic induction, $ B=\mu H $
Here, $B =$ Magnetic induction, $H =$ Intensity of magnetisation whose value $5000 A \times m$ is given.
$\mu$ is the permeability of the material of the rod.
$\therefore$ From equation (1)
$\mu=\frac{B}{H}$
$\mu=\frac{1 \text { weber }}{5000 A- m }$
$=0.0002 weber / A - m$
$=2 \times 10^{-4} A- m \quad$ Ans.
Next, $\quad \mu=\mu_0\left(1+\chi_m\right)$
or $\chi_{ m }=\frac{\mu}{\mu_0}-1$
$=\frac{0.0002}{4 \pi \times 10^{-7}}-1$
$=\frac{0.0002}{4 \times 3.14 \times 10^{-7}}-1$
$=159.23-1$ $=158.23 $ Ans.
(c) Intensity of magnetisation,
$m =\chi_{ m } \times H$
$m =158.23 \times 5000$
$=7.91 \times 10^5$ ampere $/$ meter Ans.

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Question 34 Marks

If the radius of a circular current carrying coil is doubled and the current flowing through it is reduced to half, then what will be effect on magnetic moment?

Answer

If area of current carrying coil is $A m ^2$ and current flowing through it is $i$ ampere, then magnetic moment of the coil is
$ m_1=A \times i \text { ampere } \times \text { metre }^2 $
If original radius of the coil is $r$ metre and current $i$ is flowing through it, then its magnetic moment
$m_1= A \times j=\pi r ^2 \times j$
$m_1=\pi r^2 i$
Now on doubling the radius of the coil that is on doing 2 rm ,
Area of the coil $=\pi(2 r)^2=4 \pi r^2 m^2$
On reducing the current to half its value, $ =\frac{i}{2} \text { Ampere } $
$\therefore$Now magnetic moment of the coil,
$m_2= A \times i$
$m_2=2 \pi r^2 \times \frac{i}{2}$ Ampere $\times m ^2$
$m_2=2 \pi r^2 \times i=2 \pi r^2 i$
$\therefore \quad \frac{m_2}{m_1}=\frac{2 \pi r^2 i}{\pi r^2 i}=\frac{2}{1}$
Hence, magnetic moment of the coil will become two times.

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Question 44 Marks

(a) An electron is revolving with a speed of 2 $\times 10^5 m / s$ in the orbit of radius $0.51 A$ in a hydrogen atom. Calculate the magnetic moment of the electron.
(b) Equivalent electric current due to orbital speed of electron.
(c) Magnetic field generated at the centre of the nucleus.

Answer

Given : (a) Here, radius $r=0.51 A$ $ =0.51 \times 10^{-10} m $
Speed $v=2 \times 10^5 m / s$
$\therefore$ Magnetic moment $ m=\frac{e v r}{2} $
On putting values, $ \begin{aligned} m & =\frac{\left(1.6 \times 10^{-19}\right)\left(2 \times 10^5\right)\left(051 \times 10^{-19}\right)}{2} \\ & =8.16 \times 10^{-25} Am^2 \end{aligned} $
(b) Equivalent electric current, $I =\frac{q}{l}$
Here, $q=$ magnitude of charge on electron $ =1.6 \times 10^{-19} C $ and the time taken for this charge to flow in one revolution
$t=\frac{\text { Distance travelled in one term }}{\text { Speed }}$ $=\frac{e \pi r}{v}$
$I =\frac{e}{2 \pi r / v}=\frac{e v}{2 \pi r}$
$I =\frac{\left(1.6 \times 10^{-19}\right)\left(2 \times 10^5\right)}{2 \times 3.14 \times\left(0.51 \times 10^{-10}\right)}=10^{-4} A$
(c) Magnetic field generated at the center of a circular orbit, i.e. nucleus
$B =\frac{\mu_0}{4 \pi}\left(\frac{2 \pi I }{r}\right)$
$=10^{-7}\left(\frac{2 \times 3.14 \times 10^{-4}}{051 \times 10^{-10}}\right) \frac{B}{A- m }$
$=1.23 N / A-m . $ Ans.

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Physics Case study (4 Marks)

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