Magnetic field due to a current in a solenoid - class-XI

A Toroid is a solenoid which is bent in the form of a ring with its both the ends joined together.  Since both the ends are merged, it appears endless.

A solenoid is a long coil of wire wrapped in a cylindrical shape, which creates a uniform magnetic field inside when current flows through it.

According to the Biot Savart's law, a current carrying conductor produces a magnetic field in its surroundings. This will cause the helix to oscillate across a mean position.

$\displaystyle B = \mu _0 N _0  i;   $

$B _1 = (\mu _0) \left ( \dfrac{N _0}{2} \right ) (2 i) $

$= \mu _0 N _0 i = B$

Given $d = 14 cm=0.14 m$    

Coil is unwound and is rewound into a circular coil of 4 turns,
$\therefore $ $2\pi R=4\times 2\pi r$ 
$r=\dfrac{R}{4}$
$\left | B \right | =\dfrac{\mu _{0}I}{\dfrac{2R}{4}}$ $=0.2\times$ $4=0.8T$

The length of remains same $N _1\pir=N _2\pir/2$

The torque on the rectangular coil due to presence of magnetic field is given, 

$\oint B\cdot dl=\mu _0 I _{enclosed}$
Here,
$I _{enclosed}=0$
So,  B $=0$

Magnetic dipole moment of small coil of N turns having an area A and a current $i$ is
$M = NiA$

$\vec{\tau} = \vec{M}\times \vec{B}$
$\vec{\tau} = MB \sin\theta$
$= niAB \sin\theta$          $(\because M = niA)$

$\vec{\tau} = \vec{M}\times \vec{B}$
$= ni(\vec{A}\times \vec{B})$    $(\because \vec{M} = ni\vec{A})$

According to the laws of induction of current, if the source current is increased the induced current is also increased, and the induced current is in the opposite direction to the original current.

Spring can be assumed as the coil parallel to each other.
So, when current flows in spring each coil gets current flow in same direction. So, they are attracted to each other which in turn results in compression of spring.

When a charged particle is projected at an angle $\theta$ to a magnetic field, the component of velocity parallel to the field is $v\cos \theta$ while perpendicular to the field is $v\sin \theta$, so the particle will move in a circle of radius
$r = \dfrac{m(v\sin \theta)}{qB}=\dfrac{( 1.67 \times 10^{-27}) \times \left( 4 \times 10^{5} \times \sin 60^0 \right)}{1.6 \times 10^{-19} \times 0.3}=\dfrac{ \left( 1.67 \times 10^{-27} \right) \times \left( 4 \times 10^{5} \times \dfrac{\sqrt{3}}{2} \right)}{1.6 \times 10^{-19} \times 0.3}=\dfrac{2 \times 10^{-2}}{\sqrt{3}}$
Time period: $T = \dfrac{2\pi r}{vsin\theta}= \dfrac{2\pi \times \dfrac{2 \times 10^{-2}}{\sqrt{3}}}{4 \times 10^5 \times \sin 60^0}= \dfrac{2\pi \times \dfrac{2 \times 10^{-2}}{\sqrt{3}}}{4 \times 10^5 \times \dfrac{\sqrt{3}}{2}}=\dfrac{2\pi}{3} \times 10^{-7}$
Pitch: $P = v \cos \theta T= (4\times 10^5) \times \cos 60^0 \times \dfrac{2\pi}{3} \times 10^{-7}=\dfrac{4\pi}{3}\times 10^{-2}= 4.35 \times 10^{-2}: m$

No of turns per unit length for an infinite  solenoid: $n =

\dfrac{B}{\mu _0 i}=\dfrac{3.14 \times 10^{-2}}{4\pi \times 10^{-7} \times 5}=5000$

Given that,
Number of turns in the circular coil, $N = 16$
Radius of the coil, $r = 10$ cm $= 0.1m$ 
Current in the coil, $I = 0.75A$ 
Magnetic field strength, $B = 5.0 \times 10^{2}T$ 
Frequency of oscillations of the coil, $v = 2.0s^{-1}$
Magnetic moment, $M = NIA$
$M = NI\pi r^2 = (16)(0.75)\pi(0.1)^2 = 0.3768 JT^{-1}$
The frequency is given by,
$ \nu = \dfrac{1}{2\pi}\sqrt {\dfrac{MB}{I}}$

$ I = \dfrac{MB}{4 \pi^2 \nu^2}$

$I = \dfrac{(0.3768)(5 \times 10^{2})}{4 \pi^2 2^2}$

$I = 1.19 \times 10^{4} = 1.2 \times 10^{4} Kg-m^{2}$