Magnetic Field and Magnetic Force

Magnetic field is an invisible space around a magnetic object. A magnetic field is basically used to describe the distribution of magnetic force around a magnetic object.

Magnetic fields are created or produced when the electric charge / current moves within the vicinity of the magnet. Here, the sub-atomic particle such as electrons with a negative charge moves around creating a magnetic field. These fields can originate inside the atoms of magnetic objects or within electrical conductors or wires.

Table of Content

Representation of Magnetic Field

Magnetic field can be depicted in several ways. Mathematically, it can be represented as a vector field which can be plotted as different sets on a grid. Another way is the use of field lines. The set of vectors are connected with lines. Here the magnetic field lines never cross each other and never stop.

Magnetic Field

 

Magnetic Field Unit and Measurement

The measurement of the magnetic field involves measuring its strength and direction. The measurement is necessary because every magnetic field is different from each other. Magnetic field intensity is either small and weak while some are very strong and large. If we look at the earth’s magnetic field it is weak but large.

Also Read: Electromagnetism

Nonetheless, the term magnetic field represents two unique but related fields which are usually denoted by the symbols H and B. H represents the magnetic field strength and is measured in the SI unit of ampere per meter. Meanwhile, B represents magnetic flux density and is measured in tesla.

Magnetic Force

Magnetic field depicts how a moving charge flows around a magnetic object. Magnetic force is a force that arises due to the interaction of magnetic fields. It can be either repulsive or attractive force.

Magnetic Force Acting on a Moving Charge in the Presence of Magnetic Field

A change ‘a’ is moving with a velocity ‘v’ making an angle ‘θ’ with the field direction. Experimentally, we found that a magnetic force acts on the moving charge and is given by FB=q(V×B).{{\overrightarrow{F}}_{B}}=q\left( \overrightarrow{V}\times \overrightarrow{B} \right). This is known as Lorentz force law.

Lorentz force law

Characteristics:

(1) No magnetic force acts on a stationary change present in the magnetic field.

(2) No magnetic force acts on a moving change when it is moving either parallel (or) Antiparallel to the field direction.

(3) The magnetic force acting on the moving change is maximum, when the change is moving

artoB.\bot ar\,to\,\overrightarrow{B}.

So, as VB,θ=90FB=maximum.\overrightarrow{V}\bot \overrightarrow{B},\,\theta =90{}^\circ \Rightarrow \overrightarrow{{{F}_{B}}}=\max imum.

No work is done by the magnetic force on a moving charge because FB\overrightarrow{{{F}_{B}}} is alwaysartoB\bot ar\,to\,\overrightarrow{B} .

As a result, there is no change in K.E of a change. There is no change in speed but the direction of motion may change.

Magnetic Force on Currents

Let us consider a line change ‘λ’ moving with a velocity ‘v’ as shown in the figure. The amount of change crossing a point p in a time interval Δt is λVΔt. The rate of flow of change across the point p is nothing but the line current I.

I=λVΔtΔt=λV;I=λV(Invectorform)I=\frac{\lambda V \Delta t}{\Delta t}=\lambda V\,\,\,;\,\,\overrightarrow{I}=\overrightarrow{\lambda V}(In\,vector\,form) \Rightarrow A change moving with a velocity ‘v’ (a discrete change, line change, a surface change (or) a volume change) is equivalent to the current. The force acting on a line change moving with a velocity ‘v’ (or) the magnetic force acting on a line current is given by F=L(V×B)dq=L(V×B)λd.=L(I×B)d=IL(d×B)\overrightarrow{F}=\int\limits_{L}{\left( \overrightarrow{V}\times \overrightarrow{B} \right)dq=\int\limits_{L}{\left( \overrightarrow{V}\times \overrightarrow{B} \right)\lambda d\ell .}}=\,\int\limits_{L}{\left( \overrightarrow{I}\times \overrightarrow{B} \right)d\ell }=I\int\limits_{L}{\left( \overrightarrow{d\ell }\times \overrightarrow{B} \right)}

Note: I is taken out of the integral because the currents which we are considering are study currents. A conductor PQ\overrightarrow{PQ} is carrying a current ‘I’ & is present in a magnetic field.

Magnetic Force on Currents

F=IPQd×B\overrightarrow{F}=I\int\limits_{P}^{Q}{d\overrightarrow{\ell }\times \overrightarrow{B}}

If B\overrightarrow{B} is uniform,

F=I[PQd]×B]=I[PQ×B]\overrightarrow{F}=I\left[ \int\limits_{P}^{Q}{\overrightarrow{d\ell }} \right]\left. \times \overrightarrow{B} \right]=I\left[ \overrightarrow{PQ}\times \overrightarrow{B} \right]

If it is a closed current-carrying conductor, carrying a steady current and is placed in a magnetic field B\overrightarrow{B}.

F=ILd×B\overrightarrow{F}=I\oint\limits_{L}{\overrightarrow{d\ell }}\times \overrightarrow{B}

If B\overrightarrow{B} is uniform, then, the magnetic force acting on it is always equal to zero.

Magnetic Field Of A Straight Line Current

A straight current-carrying conductor is carrying a current I as shown in the figure. By using, Biot – Savart’s law we are now finding the magnetic field at a point p, which is at a distance ‘r’ from the wire Magnetic Field Of A Straight Line Current

OA=Z cosθ=rPA\cos \theta =\frac{r}{PA}

R tanθ = Z

dz = rsec2θdθ PH=rsecθPH=r\sec \theta dB=μoIdzsinPAC4π(PA)2dB=\frac{{{\mu }_{o}}Idz\,\sin \angle PAC}{4\pi {{(PA)}^{2}}} Magnetic Field Of A Straight Line Current

=μoIrsec2θdθcosθ4π(r2sec2θ)=μoIcosθdθ4πr=\frac{{{\mu }_{o}}I\, r\sec^{2}\theta\, d\theta\, \cos \theta }{4\pi \left( r^{2}\sec^{2}\theta \right)}=\frac{{{\mu }_{o}}I\cos \theta d\theta }{4\pi r} =x[B=μoI4πrπ/2  cosθdθ=μoI4πr]x=x\left[ \overline{B}=\frac{\mu oI}{4\pi r}\int\limits_{{}}^{{\pi }/{2}\;}{\cos \theta d\theta }=\frac{{{\mu }_{o}}I}{4\pi r} \right]x =B=μoI4πrθ1θ2cosθdθ=\overrightarrow{B}=\frac{{{\mu }_{o}}I}{4\pi r}\int\limits_{-{{\theta }_{1}}}^{{{\theta }_{2}}}{\cos \theta d\theta } =μoI4πr(sin(θ1)+8n(θ2))=\frac{{{\mu }_{o}}I}{4\pi r}\left( \sin ({{\theta }_{1}})+8n({{\theta }_{2}}) \right)

 

Case I: In case of an Infinite wire carrying a current I the angle subtended by the ends at the point P can be considered θ1=θ2=π2{{\theta }_{1}}={{\theta }_{2}}=\frac{\pi }{2} B=μoI4πr(2)=μoI2πr\Rightarrow \,\overrightarrow{B}=\frac{{{\mu }_{o}}I}{4\pi r}(2)=\frac{{{\mu }_{o}}I}{2\pi r}

 

Case II: If the wire is having length extended to ∞ and the point P is present on a perpendicular passing through one end of this semi infinite wire.

Magnetic Field Of A Straight Line Current

B=μoI4πr[8n02+0]=μoI4πrθ1=0,θ2=90B=\frac{{{\mu }_{o}}I}{4\pi r}\left[ 8n{{0}_{2}}+0 \right]=\frac{{{\mu }_{o}}I}{4\pi r}\,\,\,{{\theta }_{1}}=0{}^\circ ,{{\theta }_{2}}=90{}^\circ

 

Case III: If the point p is on a ⊥ar bisector of length 2L.

B=μoI4πr(28nθ)=μoI8nθ2πr\overrightarrow{B}=\frac{{{\mu }_{o}}I}{4\pi r}\left( 28n\theta \right)=\frac{{{\mu }_{o}}I8n\theta }{2\pi r}

(or)

B=μoI2πrLL2+r2\overrightarrow{B}=\frac{{{\mu }_{o}}I}{2\pi r}\frac{L}{\sqrt{{{L}^{2}}+{{r}_{2}}}}

Magnetic Field Of A Straight Line Current

Magnetic Field of a Current-carrying Circular Loop

A circular loop carrying a current I in anti-clockwise direction lies in XY-plane with its centre at the origin. By using B.S.L, we are now calculating a magnetic field at point P.

Using B.S.L,

dB=μo4πIdsinθ(R2+z2)\overrightarrow{dB}=\frac{{{\mu }_{o}}}{4\pi }\frac{Id\ell \sin \theta }{({{R}^{2}}+{{z}^{2}})}

In case of a circular loop, dandr^arear\overrightarrow{d\ell }\,and\,\hat{r}\,are\,\bot ar

dB=μo4πid(R2+z2)\overrightarrow{dB}=\frac{{{\mu }_{o}}}{4\pi }\frac{id\ell }{({{R}^{2}}+{{z}^{2}})}

They y – components ofdB\,\overrightarrow{dB} are cancelling each other. The

B\overrightarrow{B} is contributed only by z-components

dBz{{\overrightarrow{dB}}_{z}} B=dBz=z^dBsmθ=z^μo4πIdR2+z2R(R2+z2)12\overrightarrow{B}=\oint{d{{B}_{z}}=\hat{z}}\int{\overrightarrow{dB}\,sm\theta }=\hat{z}\oint{\frac{{{\mu }_{o}}}{4\pi }\frac{Id\ell }{{{R}^{2}}+{{z}^{2}}}}\,\frac{R}{{{({{R}^{2}}+{{z}^{2}})}^{\frac{1}{2}}}} =z^μo4πIR(R2+z2)3/2d=\hat{z}\frac{{{\mu }_{o}}}{4\pi }\,\frac{IR}{{{\left( {{R}^{2}}+{{z}^{2}} \right)}^{{}^{3}/{}_{2}}}}\oint{d\ell } =z^μoIR(2πR)4π(R2+z2)3/2=z^μoI2R[1+z2R2]3/2=\hat{z}\frac{{{\mu }_{o}}IR(2\pi R)}{4\pi {{({{R}^{2}}+{{z}^{2}})}^{{}^{3}/{}_{2}}}}=\hat{z}\frac{{{\mu }_{o}}I}{2R{{\left[ 1+\frac{{{z}^{2}}}{{{R}^{2}}} \right]}^{{}^{3}/{}_{2}}}}

At the centre, z = 0, B=μoI2Rz^\overline{B}=\frac{{{\mu }_{o}}I}{2R}\hat{z}

A circular loop carrying a current I behaves as a magnetic dipole having magnetic dipole moment (m)m=I6da(m)\overrightarrow{m}=I\int\limits_{6}{da} .

In case of flat surfaces. m = I (πR2)

B=z^μom2π(R2+z2)3/2(Bofdipoleoneaxialline)\overrightarrow{B}=\hat{z}\frac{{{\mu }_{o}}m}{2\pi {{({{R}^{2}}+{{z}^{2}})}^{{}^{3}/{}_{2}}}}\left( \overrightarrow{B}\,of\,dipole\,one\,axial\,line \right)

The direction of the magnetic dipole moment depends on the sense of the current in the loop is in the anti-clockwise direction, then ‘m’ is directed ⊥ar to the plane outwards. If the current is in a clockwise direction, then the magnetic moment is directed ⊥ar to the plane inwards.

Magnetic Field Variation on the Axial Distance

At the centre, the M.F is maximum, but when we are moving to either side of the circular loop along the z-axis, the field is varying Non – linearly, at two points along the axis, the second derivative of B\overrightarrow{B} is vanishing and becoming zero. Magnetic Field Variation on the Axial Distance

The variation of B\overrightarrow{B} with Z is constant at these two-points 🡺 The variation of B\overrightarrow{B} is linear at the two points.

These two points are known as ‘the points of inflexion of the graph’.

Magnetic Field Of A Solenoid

Magnetic Field Of A Solenoid

‘N’ turns are wound on a cylindrical tube having radius ‘a’ & length ‘L’, so closely that the surface of the cylindrical tube presents a surface current density ‘k’. Assuming each turn having a circular shape.

The M.F on the axis due to each turn carrying a current I isB=μoIR22(R2+z2)3/2\overrightarrow{B}=\frac{{{\mu }_{o}}I{{R}^{2}}}{2{{({{R}^{2}}+{{z}^{2}})}^{3/2}}} . In a length dz.

The no of turns are n d z and the current I in the above formula can be replaced by ndzi of the M.F due to this current is dB=μondzia22(a2+z2)3/2\overrightarrow{dB}=\frac{{{\mu }_{o}}ndzi{{a}^{2}}}{2{{({{a}^{2}}+{{z}^{2}})}^{{}^{3}/{}_{2}}}}

z = a tan θ , dx = a secθ dθ.

dB=μonia22[asec2θdθa3[1+tan2θ]3/2]=μoniabsec2θdθ2a3sec3θ=μoni2secθdθdB=\frac{{{\mu }_{o}}ni{{a}^{2}}}{2}\left[ \frac{a{{\sec }^{2}}\theta d\theta }{{{a}^{3}}{{\left[ 1+{{\tan }^{2}}\theta \right]}^{{}^{3}/{}_{2}}}} \right]=\frac{{{\mu }_{o}}ni{{a}^{b}}{{\sec }^{2}}\theta d\theta }{2{{a}^{3}}{{\sec }^{3}}\theta }=\frac{{{\mu }_{o}}ni}{2\sec \theta }d\theta dB=μoni2cosθdθdB=\frac{{{\mu }_{o}}ni}{2}\cos \theta d\theta

This is the M.F due to the small surface current element dz of the surface current at a point p on the axis.

If the two ends of the solenoid are subtending angles θ1 and θ2 at the point p on the axis of the solenoid, then the M.F at point p is given by

B=μoniz^2θ1θ2cosθdθ\overrightarrow{B}=\frac{{{\mu }_{o}}ni\hat{z}}{2}\int\limits_{-{{\theta }_{1}}}^{{{\theta }_{2}}}{\cos \theta d\theta } =μoni2[8nθ1+8nθ2]=\frac{{{\mu }_{o}}ni}{2}\left[ 8n{{\theta }_{1}}+8n{{\theta }_{2}} \right] =μoni2[8nθ1+8nθ2]^= \frac{\mu_o ni }{2}[8n\theta_1 + 8n\theta_2 ]\hat{\not{Z}}

 

Case I: For a long solenoid at any point O inside the solenoid the M.F can be calculated as follows.

B=zaμoni2(2)=μonizn1\overrightarrow{B}=\overset{a}{\mathop{z}}\,\frac{{{\mu }_{o}}ni}{2}(2)={{\mu }_{o}}ni\overset{n}{\mathop{z}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\to 1

 

Case II: If the point P is at one end of the long solenoid.

θ1=0,θ2=π2B=znμoni22{{\theta }_{1}}=0,{{\theta }_{2}}=\frac{\pi }{2} \\\overrightarrow{B}=\overset{n}{\mathop{z}}\,\frac{{{\mu}_{o}}ni}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\to 2 \\

Comparing 1 and 2, The magnetic field is only 1/2 at the end because there is a leakage of ϕB at the ends. We can avoid this leakage of flux, by joining the ends of the solenoid. This is called endless solenoid (or) solenoid toroid.

Force and Torque on a Current Loop or Coil in a Uniform Magnetic Field

Let a rectangular current loop ABCD having length AB = CD = \ell and breadth AD = BC = b and carrying current ‘i’ be suspended in a magnetic field of flux density B. With normal (ON) to its plane making an angle Ɵ with the field direction.

Forces I (b × B) on arms AD and BC act in opposite directions along the vertical axis of suspension X Y and hence cancel.

Forces on arms AB and DC, being perpendicular to the field, are i\ellB each and they act at the middle points P and Q as shown in Fig.

Force and Torque on a Current Loop or Coil in a Uniform Magnetic Field

These forces form a couple of arm PR = PQ sin Ɵ = b sin Ɵ.

As Torque = force X arm or perpendicular distance between the two forces.

τ=iB×bsinθ\tau = i\ell B\times b \sin \theta =i(B)Bsinθ= i(\ell B) B \sin \theta =i(A)Bsinθ= i(A) B \sin \theta       (Since  b=A)\;\;\;(Since\;\ell b = A)

For a loop having n turns,

orτ=niABsinθ..(8.24)or\,\tau =n\,i\,AB\sin \,\theta \,\,\,\,\,\,\,\,\,…..\,(8.24)

If the plane of the loop makes an angle Ɵ with the direction of B. Then

τ=niABcosθ\tau =n\,i\,AB\cos \,\theta

It can be seen from equation (8.24) that the torque acting on the coil is similar to the torque acting on a magnet in a uniform magnetic field, τ=MBsinθ\,\tau =MB\sin \,\theta since the current loop behaves like a magnetic dipole of magnetic moment, M = niA.

When the plane of the circular loop is parallel to the applied magnetic field the torque becomes maximum i.e., τ=niAB.\,\tau =n\,i\,AB. When the plane of the circular loop is perpendicular to the applied magnetic field the torque becomes minimum i.e., τ=0\,\tau =0.

Couple Acting on a Bar Magnet Placed in a Uniform Magnetic Field

Let us consider a bar magnet (NS) of pole strength ‘m’ and magnetic length 2,’2\ell ‘, place in a uniform magnetic field ‘B’, as shown in the, at an angle Ɵ to the direction of the field. Each pole of the magnet is then acted upon by a force equal to mB but in opposite directions. Thus, two equal and opposite forces act on the rigid body of the magnet at the ends constituting a couple. This couple tends to rotate the magnet so as to bring it along the direction of the field.

The torque τ\tau due to the couple or the moment of the couple is equal to the product of the force and the perpendicular distance between the two forces and is given by Couple Acting on a Bar Magnet Placed in a Uniform Magnetic Field

τ=(mB)(SP)=mB(NS)sinθ{Since,  sinθ=SPNS}orτ=mB(2)sinθ{Since  NS=2}orτ=(m.2)BsinθSo,  τ=MBsinθ..(4.7)\tau =(mB)(SP)=mB(NS)\sin \theta \left\{ Since, \; \sin \theta =\frac{SP}{NS} \right\} \\ or\,\tau =mB(2\ell )\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ Since\; NS=2\ell \right\} \\ \,\,\,\,\,\,\,\,\,\,or\,\tau =(m.2\ell )B\,\sin \theta \\ \,\,\,\,\,\,\,\,\,So,\; \,\tau =MB\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…..(4.7) \\

Where M = m. (2\ell) is a constant for a given magnet and is known as the moment of the magnet. Thus the torque on the magnet depends on the angle between the magnetic field ‘B’ and the axis of the magnet. When the magnet is parallel to the direction of the field (Ɵ = 0), the torque on the magnet is equal to zero. IF the magnet is at right angles to the direction of the field (Ɵ = 90°), the torque is maximum(τmax=MB)(\tau_{max} = MB)

Magnetic Induction Due to a Bar Magnet on its Axial Line

Axial line of a magnet is the line passing through both the poles of the magnet. Let us consider a bar magnet (NS) of magnetic length ‘22\ell’and pole strength ‘m’ as shown in Fig.

Magnetic Induction Due to a Bar Magnet on its Axial Line

Let ‘A’ be a point on the axial line at a distance d from the centre (O) of the magnet. The magnetic induction at ‘A’ due to the north pole of the magnet is given by

BN=μo4πm(NA)2(RefEq.4.6)B_N = \frac{\mu_o}{4\pi}\frac{m}{(NA)^2} (Ref Eq.4.6) BN=μo4πm(d)2.(4.8)B_N = \frac{\mu_o}{4\pi}\frac{m}{(d-\ell)^{2}} ……….(4.8)

acting along NA.

Similarly, the magnetic induction at A due to the south pole of the magnet is given by

Bs=μo4πm(SA)2B_s = \frac{\mu_o}{4\pi}\frac{m}{(SA)^2} or Bs=μo4πm(d+)2acting  along  AS(4.9)B_s = \frac{\mu_o}{4\pi}\frac{m}{(d+\ell)^2} acting\;along\;AS……(4.9)

Therefore, the resultant magnetic induction at A is along NA and is given by,

BA=BNBS=μ04πm(d)2μ04πm(d+)2orBA=(μ0m4π)[1(d)21(d+)2]=μ04πm.4d(d22)2{{B}_{A}}={{B}_{N}}-{{B}_{S}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{(d-\ell )}^{2}}}-\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{(d+\ell )}^{2}}}\,\,\,\,\,\,\, \\ \,\,\,\,\,\,\,\,\,\,\,\,or\,{{B}_{A}}=\left( \frac{{{\mu }_{0}}m}{4\pi } \right)\left[ \frac{1}{{{(d-\ell )}^{2}}}-\frac{1}{{{(d+\ell )}^{2}}} \right]=\frac{\mu 0}{4\pi }\frac{m.4d\ell }{{{({{d}^{2}}-{{\ell }^{2}})}^{2}}} \\ orBA=μ04πm(2)2d(d22)2=μ04π2MD(d22)2(Since,  M=m2)or\,{{B}_{A}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m(2\ell )2d}{{{\left( {{d}^{2}}-{{\ell }^{2}} \right)}^{2}}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2MD}{{{({{d}^{2}}-{{\ell }^{2}})}^{2}}}\,\,\,\,\,\,\,\,\,\,(Since, \; M=m2\ell )

Where M = m (22\ell) is the magnetic moment of the magnet.

Thus the magnetic induction at any point on the axial line of a bar magnet is given by

BA=μ04π2Md(d22)2..(4.10)\,{{B}_{A}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2Md}{{{({{d}^{2}}-{{\ell}^{2}})}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…..(4.10)

The direction of the magnetic induction BA on the axial line is always in the same direction as the magnetic moment.

In the case of a short bar magnet, i.e., when \ell < < d, we can approximate the Eq 4.10 as

BA=μ04π2Md3..(4.11){{B}_{A}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{d}^{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,……..(4.11)

Magnetic Induction Due to a Bar Magnet on its Equatorial Line

Magnetic Induction Due to a Bar Magnet on its Equatorial Line

Equatorial line of a magnet is the perpendicular bisector of the axis of the magnet. Let us consider a point E on the equatorial line of a bar magnet (NS) at a distance d from the centre ‘O’ of the magnet. Let ‘m’ be the pole strength and ‘22\ell’ be the distance between the two poles of the magnet.

The magnetic induction at ‘E’ due to the north pole of the magnet is given by

BN=μ04πm(NE)2=μ04πm(d2+2)alongNE{{B}_{N}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{(NE)}^{2}}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{({{d}^{2}}+{{\ell }^{2}})}along\,NE

 

Similarly, the magnetic induction at ‘E’ due to the south pole of the magnet can be written as

BS=μ04πm(SE)2=μ04πm(d2+2)alongES{{B}_{S}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{(SE)}^{2}}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{({{d}^{2}}+{{\ell }^{2}})}along\,ES

BN and BS can be vectorially represented along (EP and EQ) the sides of the parallelogram EPRQ. Then ER the diagonal of the parallelogram gives the resultant field induction BE. From the similar triangle EPR and NES we have

ERNS=EPNEOrER=EPNE.(NS)=BN2d2+2\frac{ER}{NS}=\frac{EP}{NE}Or\,ER=\frac{EP}{NE}.(NS)=\frac{BN\,2\ell }{\sqrt{{{d}^{2}}+{{\ell }^{2}}}}

ER=μ04πm.2(d2+)d2+2ER=\frac{{{\mu }_{0}}}{4\pi }\frac{m.2\ell }{({{d}^{2}}+\ell )\sqrt{{{d}^{2}}+{{\ell }^{2}}}} orBE=μ04πM(d2+2)3/2.(4.12)or\,{{B}_{E}}=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{\left( {{d}^{2}}+{{\ell }^{2}} \right)}^{3/2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,….(4.12)

The direction of the field induction BE on the equatorial line is always opposite to the direction of the magnetic moment. In the case of a short bar magnet, i.e., when \ell < < d, we can approximate Eq. 4.12 as

BE=μ04πMd3.(4.13){{B}_{E}}=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{d}^{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,….(4.13)