Q: The dimension of \$1/2\varepsilon_oE^2\$(\$\varepsilon_o\$ permittivity of free space; E: electric field) is?

Dimension of \$E_0=M^{-1}L^{-3}T^4A^2\$Dimension of E(Electric field)=\$MLT^{-2}/AT\\=MLT^{-1}A^{-1}\$Dimension of \$\cfrac{1}{2\varepsilon _0 E^2}=\cfrac{1}{(M^{-1}L^{-3}T^4A^2)(MLT^{-1}A^{-1})^2}\\=\cfrac{1}{ML^{-1}T^2}=ML^{-1}T^{-2}\$

Q: State and explain Biot-Savart Law.

Hint: The Biot-Savart law states how the value of the magnetic field at a specific point in space from one short segment of current-carrying conductor depends on each factor that influences the field.Solution:Step1: Application of Biot-Savart lawThe Biot-Savart Law is used to determine the magnetic field intensity B near a current carrying conductor in terms of magnetic field intensity generated by its source current element. Consider a wire carrying an electric current I and also consider an infinitely small length of a wire dl at a distance r from point AStep2: Statement of Biot-Savart lawIt is clear from above free body diagram that, Biot-Savart Law states that magnetic intensity dB at a point A due to current I flowing through a small element dl is,a) Directly proportional to current (I). b) Directly proportional to the length of the element (dl). c) Directly proportional to the sine of angle θ between the direction of current and the line joining the element dl from point A. d) Inversely proportional to the square of the distance (x) of point A from the element dl. \$dB\alpha \dfrac{{Idl\sin \theta }}{{{r^2}}}\$\$dB = K\dfrac{{Idl\sin \theta }}{{{r^2}}}\$ Where, \$K = \dfrac{{{\mu _0}{\mu _r}}}{{4\pi }}\$ μ0= Absolute permittivity of air in vacuum μr= relative permittivity.

Q: A toroid has a core (non -ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

\$r_1=0.25\ m, r_2=0.26\ m, n=3500, I=11\ A\$(a)magnetic field outside is zero. It is non zero only inside the core of a toroid.By definition, Ampere's Law states that integration of magnetic field over an amperian loop is equivalent to constant μ multiplied by current enclosed. Now, the current that flows at outer circumference and inner circumference of toroid are in opposite direction to each other. Therefore, current enclosed by Amperian loop is zero. Thus, magnetic field outside the toroid is zero.(b)Magnetic field inside the core of a toroid is given by the relation \$B=\dfrac{\mu_onI}{l}\$l=length of toroidwhere, \$l=2\dfrac{\pi(r_1+r_2)}{2}=.51\pi\$ .\$B=\dfrac{4\pi\times 10^{-7}\times 3500\times 11}{0.51\pi}\$so, \$B=3\times 10^{-2}\ T\$ (c)Magnetic field in the empty space surrounded by the toroid is zero as there will no current in the empty space.

Question 1

Derive the expression for magnetic field at a point on the axis of a circular current carrying loop.

Accepted Answer

In the above diagram,Let O be the centre of the ring, R be the radius of the ring. The point P lie outside the ring on the axis, let the distance between the P and o be $x$, where we have to calculate the magnetic field.Let $\vec{dB}$ be the magnetic field due to a small length $dl$ of the ring,As per bio-severt rule,$\vec{dB}=\dfrac{\mu_o Idl}{4\pi r^2} $We know that $r=\sqrt{(x^2+R^2)}$So $\vec{db}=\dfrac{\mu_o I dl}{4\pi \sqrt{x^2+R^2}}$This field have vertical$\vec{dB}_y=\vec{dB}sin\theta$ and horizontal component$\vec{dB}_x=\vec{dB} cos\theta$,Vertical component will cancel out each other, only horizontal components are responsible, So, $\vec{B}=\vec{dB}_1+\vec{dB}_2+.....$$\vec{dB}=\dfrac{\mu_o I dl sin\theta}{4\pi \sqrt{R^2+x^2}}$$\vec{dB}=\dfrac{\mu_o I x}{4\pi (R^2+x^2)^{\dfrac{3}{2}}}$The magnetic field due to the circular current loop of radius a at a point which is a distance R away, and is on its axis,So $\vec{B}=\dfrac{\mu_o I x^2}{2(R^2+x^2)^{\dfrac{3}{2}}}$

Question 2

Derive an expression of the magnetic field at the centre of a circular current carrying coil.

Accepted Answer

Consider a circular coil of radius a and carrying current I in the direction shown in Figure. Suppose the loop lies in the plane of paper. It is desired to find the magnetic field at the centre O of the coil. Suppose the entire circular coil is divided into a large number of current elements, each of length dl. According to Biot-Savart law, the magnetic field  $\overrightarrow{dB}$  at the centre O of the coil due to current element $I\overrightarrow{dl}$ is given by, $\overrightarrow{dB}=\dfrac{\mu_o I (\overrightarrow{dl}\times \overrightarrow{r})}{4\pi r^3}$ where  $\overrightarrow{r}$  is the position vector of point O from the current element. The magnitude of  $\overrightarrow{dB}$  at the centre O is ${dB}=\dfrac{\mu_o I {dl}a sin\theta}{4\pi a^3}$ $\therefore {dB}=\dfrac{\mu_o I {dl}sin\theta}{4\pi a^2}$ The direction of $\overrightarrow{dB}$  is perpendicular to the plane of the coil and is directed inwards. Since each current element contributes to the magnetic field in the same direction, the total magnetic field B at the center O can be found by integrating the above equation around the loop i.e.$\therefore B=\lmoustache{dB}=\lmoustache\dfrac{\mu_o I {dl}sin\theta}{4\pi a^2}$ For each current element, angle between  $\overrightarrow{dl}$ and  $\overrightarrow{r}$  is 90&#176;. Also distance of each current element from the center O is a.$\therefore B=\dfrac{\mu_o I sin90^o }{4\pi a^2}\lmoustache dl$ But $\lmoustache dl= 2\pi a=$total length of the coil$\therefore B=\dfrac{\mu_o I  }{4\pi a^2}2\pi a$ $\therefore B=\dfrac{\mu_o I }{2a}$

Question 3

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic filed at the centre of the loop is B. It is them bent into circular coil of n turns. The magnetic field at the centre of this cold of n turns will be.

Accepted Answer

The circumference of the first loop is the length of the wire=$2\pi R$The magnetic field at the center of this wire=$B=\dfrac{\mu_0I}{2R}$The same wire is bent into n circular coils.Thus new radius is found out by$n\times 2\pi r=2\pi R$$\implies r=\dfrac{R}{n}$Thus magnetic field at the center of these loops=$n\dfrac{\mu_0 I}{2r}$$=\dfrac{n^2\mu_0I}{2R}=n^2B$

Question 4

The dimension of $1/2\varepsilon_oE^2$($\varepsilon_o$ permittivity of free space; E: electric field) is?

Accepted Answer

Dimension of $E_0=M^{-1}L^{-3}T^4A^2$Dimension of E(Electric field)=$MLT^{-2}/AT\=MLT^{-1}A^{-1}$Dimension of $\cfrac{1}{2\varepsilon _0 E^2}=\cfrac{1}{(M^{-1}L^{-3}T^4A^2)(MLT^{-1}A^{-1})^2}\=\cfrac{1}{ML^{-1}T^2}=ML^{-1}T^{-2}$

Question 5

State and explain Biot-Savart Law.

Accepted Answer

Hint: The&#10;Biot-Savart law states how the value of the magnetic field at a specific&#10;point in space from one short segment of current-carrying&#10;conductor depends on each factor that influences the field.Solution:Step1: Application of Biot-Savart lawThe Biot-Savart&#10;Law is used to determine the magnetic field intensity B near a&#10;current carrying conductor in terms of magnetic field intensity generated by&#10;its source current element. Consider a wire carrying an electric&#10;current I and also consider an infinitely small length of a wire dl&#10;at a distance r from point AStep2: Statement of&#10;Biot-Savart lawIt is clear from above&#10;free body diagram that, Biot-Savart  Law states that magnetic&#10;intensity dB at a point A due to current I flowing through&#10;a small element dl is,a) Directly proportional&#10;to current (I).&#10;b) Directly proportional&#10;to the length of the element (dl).&#10;c) Directly proportional&#10;to the sine of angle &#952; between the direction of current and the line joining&#10;the element dl from point A.&#10;d) Inversely&#10;proportional to the square of the distance (x) of point A from the&#10;element dl.&#10;$dB\alpha \dfrac{{Idl\sin \theta }}{{{r^2}}}$$dB = K\dfrac{{Idl\sin \theta }}{{{r^2}}}$&#10;Where, $K = \dfrac{{{\mu _0}{\mu _r}}}{{4\pi }}$    &#956;0&#8203;= Absolute&#10;permittivity of air in vacuum     &#956;r&#8203;= relative&#10;permittivity.&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;

Question 6

What is a solenoid? Draw the pattern of magnetic field lines of (i) A current carrying solenoid and (ii)A bar magnet. List two distinguishing features between the two fields.

Accepted Answer

It is coil containing many circular terns, there wire are wrapped closely in the shape of a cylinder and it is used as an electromagnet.It also refers to any device that converts electrical energy to mechanical energy using a solenoid. The device creates a magnetic field from electric current and uses the magnetic field to create linear motion.In above image:-1). The magnetic field for the current carrying solenoid 2). For bar magnet.List of the distinguishing features between the two fields:a) The poles of the bar magnet do not lie exactly at the end of the magnet but are somewhat inside. In a solenoid, poles can be considered to be lying at the edge.b) The magnetism retains in the bar magnet naturally but in the solenoid, the magnetism is there so long current flows through it.c) A magnetic field of a bar magnet emanates from throughout the body of the magnet, with more intensity at the poles. While there is no field emanating from the lateral surface of the solenoid.

Question 7

Write Ampere's circuital law.Obtain an expression for magnetic field on the axis of current carrying very long solenoid. Draw necessary diagram.

Accepted Answer

Ampere's circuital law states that line integral of magnetic field around any closed loop is equal to $\mu_o$ times the electric current flowing through the cross-section area enclosed by that loop.Mathematically,   $\oint B.dl  = \mu I$Let the current flowing in the solenoid having number of turns per unit length $n$ be $I$.Magnitude of magnetic field inside the solenoid is $B$ while at outside is zero.Now   $\oint_{loop} B.dl  = \int B_{ab}. L  + \int B_{bc}. L'  + \int B_{cd}. L + \int B_{da} . L'$The value of first term   $\int B_{ab}.L = BL$The second and fourth term are zero because angle between magnetic field and the length loop is $90^o$.The third term is also zero as the value of magnetic field outside the solenoid is zero.Total current flowing through the loop   $I_{total} = (nL) I $From Ampere's circuital law, we get    $BL = \mu_o (nL I)$ $\implies$   $B = \mu_o nI$

Question 8

A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is in which direction?

Accepted Answer

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Question 9

A toroid has a core (non -ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Accepted Answer

$r_1=0.25\ m, r_2=0.26\ m, n=3500, I=11\ A$(a)magnetic field outside is zero. It is non zero only inside the core of a toroid.By definition, Ampere's Law states that integration of magnetic field over an amperian loop is equivalent to constant &#956; multiplied by current enclosed. Now, the current that flows at outer circumference and inner circumference of toroid are in opposite direction to each other. Therefore, current enclosed by Amperian loop is zero. Thus, magnetic field outside the toroid is zero.(b)Magnetic field inside the core of a toroid is given by the relation $B=\dfrac{\mu_onI}{l}$l=length of toroidwhere, $l=2\dfrac{\pi(r_1+r_2)}{2}=.51\pi$ .$B=\dfrac{4\pi\times 10^{-7}\times 3500\times 11}{0.51\pi}$so, $B=3\times 10^{-2}\ T$ (c)Magnetic field in the empty space surrounded by the toroid is zero as there will no current in the empty space.

Question 10

Draw a labelled diagram showing the magnetic field lines of a loop carrying current. Mark the direction of current and the direction of magnetic field by arrows in your diagram.

Accepted Answer

Concentric circles are formed, which are centered at the point on both sides of the loop as shown in figure. The lines near the center of the loop are almost straight. The magnetic field at the center of the loop is perpendicular to the plane of the loop. The concentric circles become larger as we move away from the wire because as the distance increases from the current carrying conductor the strength of the magnetic field fades away.