4 Marks Questions

Question 14 Marks

Derive an expression for the magnetic field at any point on the axis of a current carrying loop from Bio-Savart's law. Draw the necessry fig.

Answer

When the observation point is located at the centre of the coil-we consider a circular coil whose radius is $a$, in which current I is flowing. When the observation point is located at the center of the coil, then the distance of the observation point from each part of the coil $a$ as shown in the figure is equal to the radius $r$ of the coil. Apart from this, the position vector $P$ relative to each current lop makes an angle of $90^{\circ}$ with the direction of the current due to which $\sin \theta=\sin 90^{\circ}=1$.

Hence, according to Biot-Savart law, the magnetic field at the observation point due to loop of the coil is
$ \delta B =\frac{\mu_0}{4 \pi} \frac{I / l}{a^2} $
Due to all the loops, the direction of the magnetic field is perpendicular to the plane of the coil as per the clockwise screw rule. Hence, the total magnetic field at the center of the coil due to complete coil is
$B =\Sigma \delta B =\frac{\mu_0}{4 \pi} \frac{1}{a^2} \Sigma \delta l$
If the coil has $n$ turns of equal radius then
$=\frac{\mu_0}{4 \pi} \cdot \frac{l}{a^2}(2 \pi a)=\frac{\mu_0 I }{2 a}$
$\delta l=2 \pi n a$
$B =\frac{\mu_0 n I }{2 a}$ Tesla
The direction of the magnetic field $B$ is perpendicular to the plane of the coil and this direction is known by the right hand palm rule. If in the coil of 1 metre radius having 1 turn, 1 ampere current is passed then the value of mgnetic induction at the centre of the coil will be as follows:
$B =\frac{4 \pi \times 10^{-7} \times 1 \times 1}{2 \times 1}$
$=2 \pi \times 10^{-7}$ Tesla.
Therefore, 1 ampere is the current which, when flows in a coil of 1 metre radius having 1 turn produces $2 \times 10^{-7}$ Weber $/ m ^2$ (Tesla) magnetic induction is produced at the centre of the coil.

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Question 24 Marks

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of magnetic field due to the current 1.5 m in below the line?

Answer

Given Current from east to west, I = 90A
$r=1.5 m$

$B =\frac{\mu_0 I }{2 \pi r}$
Putting the values,
$B=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}=\frac{180 \times 10^{-7}}{1.5}$
$=1.2 \times 10^{-5}$
Direction of B: By right hand thumb rule, direction of magnetic field at the observation point is towards south direction.

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Question 34 Marks

(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of $60^{\circ}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Answer

Given : Number of turns in the circular coil,
$N =30$
Radius of the coil, $r=8.0 cm$
$=8 \times 10^{-2} m$
Current in the coil, $I=6.0 A$
Uniform magnetic field,
$B=1.0 T$
$\theta=60^{\circ}$
Area $A =\pi r^2$
$=3.14 \times\left(8 \times 10^{-2}\right)^2$
$=3.14 \times 64 \times 10^{-4}$
$=200.96 \times 10^{-4}$
$\cong 2.01 \times 10^{-2} m^2$
(a) Magnitude of torque acting on current carrying coil due to magnetic field,
$\tau=\text{ NIBA} \sin \theta$
On putting the values,
$\tau=30 \times 6 \times 1 \times 2.01 \times 10^{-2} \times \sin 60^{\circ}$
$=180 \times 2.01 \times 10^{-2} \times \frac{\sqrt{3}}{2}$
$=3.1\text{ Nm}$
To stop the coil from rotating, we need to apply torsion equal and opposite to torque $\tau$
$\therefore$ Desired counter torque
$=\tau= 3.1\text{ Nm}$
(b) No, the answer does not change because the formula$\tau$=NIAB sin $\theta$ is true for plane loop of any size.

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Question 44 Marks

In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^-4T ) is maintained. An electron is shot into the field with a speed of 4.8 X 10⁶m s^-1normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
$\left( e =1.6 \times 10^{-19} C , m_e=9.1 \times 10^{-31} kg\right)$

Answer

Given :
$B=6.5 G=6.5 \times 10^{-4} T$
$v=4.8 \times 10^6 m / s$
r = ?
$\begin{aligned} e & =1.6 \times 10^{-19} C
\\m_e & =9.1 \times 10^{-31} kg .\end{aligned}$
If radius of the circular orbit is r, then the force B due to magnetic field on moving electron is
$F =e v B \sin \theta$
Direction of force is perpendicular to both v and B. Hence electron will move in a circular path. If radius of the path covered by the electron is r, then
$e v B \sin 90^{\circ}=\frac{m_e v^2}{r}$
$\Rightarrow \quad e v B=\frac{m_e v^2}{r} \quad\left[\because \sin 90^{\circ}=1\right]$
or$r=\frac{m_e v}{e B}$
On putting the values,
$r=\frac{9.1 \times 10^{-31} kg \times 4.8 \times 10^6 m / s}{1.6 \times 10^{-19} C\times 6.5 \times 10^{-4} T}$
$\begin{array}{l}=4.2 \times 10^{-2} m \\
=4.2 cm \text { Ans. }\end{array}$

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