Magnetism Concept Page - 9

Definition

Analogy between electric dipole and magnetic dipole

Electric field due to an electric dipole is given by:
E=p4πεor31+3cos2θ−−−−−−−−−√$E={\displaystyle \frac{p}{4\pi {\epsilon }_o{r}^3}}\sqrt{1+3{\cos }^2\theta }$
p:$p:$ Electric Dipole Moment

Magnetic field due to a magnetic dipole is given by:
B=μoM4πr31+3cos2θ−−−−−−−−−√$B={\displaystyle \frac{{\mu }_{o}M}{4\pi r^3}}\sqrt{1+3{\cos }^2\theta }$

There is a clear analogy between the electric field and the magnetic field of electric and magnetic dipoles respectively.

Definition

Solenoid

A solenoid is a long wire wound in the form of a helix where the neighboring turns are closely spaced. So each turn is regarded as a circular loop. The net magnetic field is the vector sum of the fields due to all the turns. Enamelled wires are used for winding so that turns are insulated from each other. Generally, the length of the solenoid is large as compared to the transverse dimension. For example, if the solenoid has circular turns, the length is large as compared to its radius. If it has rectangular turns, the length should be as large as compared to its edges.

Definition

Terrestrial Magnetism

Our earth behave in such a way that we can assume the existence of a very strong magnet inside it. The magnetism of earth is called terrestrial magnetism.

Result

Magnetic field due to a solenoid

Solenoid is same as a bar electromagnet. Its magnetic field lines are shown in the diagram.

Definition

Magnetic field due to a finite solenoid

Consider a rectangular Amperian loop abcd. Along cd the field is zero. Along transverse sections bc and ad, the field component is zero. Thus, these two sections make no contribution. Let the field along ab be B$B$. Thus, the relevant length of the Amperian loop is, L=h$L=h$. 
Let n$n$ be the number of turns per unit length, then the total number of turns is nh$nh$. The enclosed current is, Ie=I(nh)$I_e=I(nh)$ , where I$I$ is the current in the solenoid. From Amperes circuital law
BL=μ0Ie$BL={\mu }_0{I}_e$, Bh=μ0I(nh)$Bh={\mu }_{0}I(nh)$
B=μ0nI$B={\mu }_{0}nI$ 
The direction of the field is given by the right-hand rule

Example

Axial magnetic field of a finite solenoid

A single-layer coil (solenoid) has length l$l$ and cross-section radius R$R$. A number of turns per unit length is equal to n$n$. The magnetic induction at the centre of the coil when a current I$I$ flows through it is given as B=xμ0nI1+(2Rl)2−−−−−−−−−−√${\displaystyle B=\frac{x{\mu }_{0}nI}{{\displaystyle \sqrt{1+{\left(\frac{2R}{l}\right)}^2}}}}$. Find x$x$.Magnetic field at the centre of a finite solenoid , Bcentre=μonI2[cosθ1+cosθ2]$B_{centre}={\displaystyle \frac{{\mu }_{o}nI}{2}}[cos{\theta }_1+cos{\theta }_2]$ Here, θ1=θ2=θ${\theta }_1={\theta }_2=\theta$⟹$⟹$ Bcentre=μonIcosθ$B_{centre}={\mu }_{o}nIcos\theta$Now cosθ=l/2R2+(l2)2−−−−−−−−√=11+(2Rl)2−−−−−−−−−√$cos\theta ={\displaystyle \frac{l/2}{\sqrt{R^2+({\displaystyle \frac{l}{2}}{)}^2}}}={\displaystyle \frac{1}{\sqrt{1+({\displaystyle \frac{2R}{l}}{)}^2}}}$
Thus, Bcentre=μonI1+(2Rl)2−−−−−−−−−√$B_{centre}={\displaystyle \frac{{\mu }_{o}nI}{\sqrt{1+({\displaystyle \frac{2R}{l}}{)}^2}}}$
Hence, x=1$x=1$

Definition

Magnetic fields of finite and infinite solenoid

The magnetic field lines of a finite solenoid form continuous closed loops, unlike the electric dipole.In the infinite solenoid, the lines of the magnetic fields are all parallel to its axis and do not come out from the coil.
An infinite solenoid is given by the superposition of two facing semi-infinite solenoids.The field inside an infinite solenoid is constant and parallel to its axis and does not vary across the radius of the solenoid.

Definition

Toroid

If a solenoid is bent in a circular shape and the ends are joined, we get a toroid. Alternatively, one can start with a nonconducting ring and wind a conducting wire closely on it. The magnetic field in such a toroid can be obtained using Ampere's Law.

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