3 Marks Question

Question 13 Marks

"Write ampere's circuital law. In a long straight wire of cross-sectional cut (radius a), a direct current is flowing in the wire. Current is uniformly distributed in the wire. Calculate the magnetic field in inner surface of the wire ( $r

Answer

According to Ampere's law, the line integral of the magnetic field around a closed path is equal to the product of magnetic permeability of that space and the algebraic sum of the current passing through that closed path in vacuum or air.
In mathematical form,
$ \oint \overrightarrow{B} \cdot \overrightarrow{d l}=\mu_0 \Sigma I $
where, $\mu_0=$ magnetic permeability of vacuum and $\oint \overrightarrow{ B } \cdot \overrightarrow{d l}=$ is known as line integral of magnetic field $(\overrightarrow{ B })$.
We need to calculate the magnetic field at $r<a$ and for this ampere loop is that circle on which I is marked. On taking radius of the circle to be r for this loop,

Now here, the electric current associated is not I but less than this. Since electric current is distributed uniformly, therefore, magnitude of element of associated electric current,
$ I_e=\left(\frac{\pi r^2}{\pi a^2}\right)=\frac{l r^2}{a^2} $
On applying Ampere's law
$B \times(2 \pi r)=\frac{\mu_0 I r^2}{a^2}$
$\Rightarrow \quad B =\left(\frac{\mu_0 I r^2}{a^2}\right) \times \frac{1}{2 \pi r}$
$B =\left(\frac{\mu_0 I }{2 \pi a^2}\right) r$

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Question 23 Marks

Current of 4 A is flowing in a straight wire AB. A proton $P$ is travelling parallel to the wire in the opposite direction of current with a speed of $4 \times 10^6$ $m / s$ at a distance of 0.2 m from the wire as shown in the diagram. Calculate the magnitude of force applied on the proton and also tells its direction.

Answer

Magnetic field produced in long wire located in plane of paper at a distance of 0.2 m
$ \begin{aligned} B & =\frac{\mu_0}{4 \pi}\left(\frac{2 I}{r}\right)=10^{-7}\left(\frac{2 \times 4}{0.2}\right) \\ & =4 \times 10^{-6} \text { weber } / m^2 \end{aligned} $

Due to second rule of righthand palm rule, the direction of $\vec{B}$ will be perpendicular to the plane of paper inwards. Proton is moving perpendicular to it.
Magnitude of force acting on it $ F=q v B \sin 90^{\circ}=q v B $
On substituting the values,
$F=1.6 \times 10^{-19} \times\left(4 \times 10^6\right) \times 4 \times 10^{-6}$
$=1.6 \times 16 \times 10^{-19}$
$=25.6 \times 10^{-19}=2.56 \times 10^{-18} N$
According to Fleming's left-hand rule, direction of force on paper will be in plane of paper, perpendicular to the wire, away from it which means towards the right.

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Question 33 Marks

Two long parallel wires are 8 cm apart. Currents of magnitudes $i$ and $3 i$ are flowing in them respectively in the same direction. Where the resultant magnetic field due to two will be zero?

Answer

Sol. Let magnetic field at distance $x$ from wire carrying $i$ current, due to two wires is zero, then distance from the wire carrying $3 i$ current is $(8-x) cm$. Therefore, magnetic field at distance $x$ due to wire carrying $i$ current
$B _1=\frac{\mu_0 i}{2 \pi x}$
Magnetic field due to wire carrying $3 i$ current at a distance $(8-x)$
$B _2=\frac{\mu_0(3 i)}{2 \pi(8-x)}$
According to the question,
$B _1- B _2=0$ or $B _1= B _2$
From equations (1) and (2)
$\frac{\mu_0 i}{2 \pi x}=\frac{\mu_0(3 i)}{2 \pi(8-x)}$
or $\frac{1}{x}=\frac{3}{8-x}$
or $8-x=3 x$
or $8=4 x$
or $x=\frac{8}{4}=2 cm$
Hence, at a distance of 2 cm from the wire carrying current $i$, the resultant magnetic field due to two wires will be zero.

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Question 43 Marks

An alpha particle enters a magnetic field of 0.2 weber $/ m ^2$ with a speed of $6.0 \times 10^5 m / s$ perpendicular to the field. Find out the acceleration of the particle and radius of the path.

Answer

Given : For Alpha particle
$\therefore \quad q=2 e=2 \times 1.6 \times 10^{-19} C$
$=3.2 \times 10^{-19} C$
$m =4 \times 1.67 \times 10^{-27} kg$
$=6.68 \times 10^{-27} kg$
$B =0.2$ weber $/ m ^2$
$v=6.0 \times 10^5 m / s$
$\theta=90^{\circ}$
Force on alpha particle due to magnetic field $F=q v B \sin \theta $
On putting the values
$=3.2 \times 10^{-19} \times 6.0 \times 10^5 \times 0.2 \times \sin 90^{\circ}$
$=3.2 \times 1.2 \times 10^{-14} \times 1$
$=3.84 \times 10^{-14} N$
Acceleration of the particle,
$a=\frac{ F }{m}=\frac{3.84 \times 10^{-14}}{6.68 \times 10^{-27}}$
$=5.75 \times 10^{12} m / s ^2$
Radius of the path of alpha particle,
$r=\frac{m \nu}{q B}$
$r=\frac{6.68 \times 10^{-27} \times 6.0 \times 10^5}{3.2 \times 10^{-19} \times 0.2}$
$=\frac{6.68 \times 6 \times 10^{-22}}{0.64 \times 10^{-19}}$
$=6.3 \times 10^{-2}$ meter Ans.

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Question 53 Marks

Write Biot-Savart law. Write the path of the motion of an electron when it enters magnetic field (a) perpendicularly (b) at $\theta$ angle.

Answer

Biot-Savart's law : When current flows through a conductor then a magnetic field is produced around it. To find out the intensity of magnetic field at any point, Biot and Savart propounded a law based on their experiments which is known as Biot-Savart law. According to this rule, due to a small part $\delta /$ of any current carrying conductor, magnitude of magnetic induction or intensity of magnetic field $\delta /$ at a point $P$ is
(i) proportional to the magnitude of current I.
(ii) proportional to the current fraction length $\delta l$.
(iii) The angle between the position vector $\vec{r}$ of the observation point relative to the current fraction $\overrightarrow{\delta l}$ is directly proportional to the $\sin \theta$ of the $\theta$ i.e. $\sin \theta$.
(iv) inversely proportional to the square of distance $r$ from the observation point to the element length.
$ \text { i.e. } \quad \delta B \propto \frac{I \delta / \sin \theta}{r^2} $
In vacuum or magnetic medium
$ \delta B=\left(\frac{\mu_0}{4 \pi}\right) \frac{I \delta / \sin \theta}{r^2} $
Where $\mu_0=4 \pi \times 10^{-7}$ weber/ampere or henry/meter and is known as magnetic permeability of vacuum $\mu_0$.
(a) Circular,
(b) In the shape of coil.

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