Magnetic force experienced by a moving charged particle inside uniform magnetic field:
Let us consider, a charged particle having magnitude +q moves along XY plane with velocity v by making an angle [latex]\theta[/latex] with B. B represent magnitude of the magnetic field strength produced by the magnetic field. A magnetic force is act in a perpendicular direction of the plane contained by v and B. Fm represent magnetic force experienced by charged particle along z-axis. Experimentally, it is found that the magnitude of the magnetic force experienced by charged particle is:
i. directly proportional to magnitude of a charged particle i.e.
F ∝ q.
ii. directly proportional to magnetic field strength i.e.
F ∝ B.
iii. directly proportional to velocity of the charged particle i.e.
F ∝ v.
iv. directly proportional to sine of angle between v and B. i.e.
F ∝ Sin[latex]\theta[/latex].
Combing these four relations, we get,
F ∝ BqvSin[latex]\theta[/latex]
Or, F = kBqvSin[latex]\theta[/latex], where, k = Proportionality constant = value 1
[latex]\therefore F = Bqvsin\theta[/latex] ……….. (v)
Hence, eqn. (v) represents the expression for force experienced by charged particle inside a magnetic field.
In vector form, above expression can be written as:
[latex]\vec{F_m} = q(\vec{v}\times \vec{B})[/latex]
Which is called Lorentz magnetic force. The direction of the Lorentz force explained by Fleming’s left-hand rule which states that, “when three finger of left-hand side are mutually perpendicular to each other, thumbs shows the direction of magnetic force, index finger shows the direction of magnetic field strength and middle finger shows the direction of motion of charged particle.
Special Cases
The magnitude of the force experienced by charge particle when it moves inside the magnetic field is given by:
Fm = [latex]Bqvsin\theta[/latex]
(i) When charge particle moves parallelly or anti-parallelly with the direction of the magnetic field, we get,
Fm = BqvSin0o
Or, BqvSin180o.
[latex]\therefore F_m = 0[/latex], which is called minimum force.
(ii) When charge particle moves perpendicularly with the direction of field, we get,
Fm = BqvSin90o
= Bqv
(iii) When charged particle are at rest inside the magnetic field, v = 0. Then,
Fm = [latex]Bq.0.sin\theta[/latex]
Fm = 0.
(iv) When charged particle are electrically neutral, q = 0, then,
Fm = [latex]B.0.v.sin\theta[/latex]
[latex]\therefore[/latex] Fm = 0.
Magnetic force experienced by the current carrying conductor:
Let us consider, a current carrying conductor having length ‘L’ and cross sectional area ‘A’ which is kept inside a uniform magnetic field strength B. The conductor inclined in the direction of the magnetic field by an angle [latex]\theta[/latex]. I represent steady amount of electric current which is passed through the conductor ‘vd’ represent drift velocity of the electrons inside the conductor, ‘n’ represents total number of free electrons per unit volume and ‘e’ represent electronic charge.
The magnetic force experienced by a single electron inside a uniform magnetic field becomes Fe. i.e.
Fe = [latex]Bev_d sin\theta[/latex]
The total no. of free electrons contained by the current carrying conductor becomes,
N = nAL
Now, the total magnetic force experienced by the current carrying conductor inside a field becomes,
Fm = N.Fe
= [latex]Bev_d sin\theta.nAL[/latex]
= (nevdA) [latex]BLsin\theta[/latex]
[latex]\therefore F_m = BILsin\theta[/latex] ………… (i)
Where, I = vdenA.
Hence, eqn. (i) represents the expression for magnitude of magnetic force experienced by current carrying conductor. Mathematically, it can be written as,
[latex]\vec{F_m} = I(\vec{L}\times \vec{B})[/latex]
It is explained by Fleming left hand rule which state that when 3 fingers of the left hand side are mutually perpendicular to each other. Middle finger shows the direction of induced current; index finger shows the direction of magnetic field and thumb shows the direction of magnetic force.
Special Case:
The magnitude of the magnetic force experienced by current carrying conductor is given by:
Fm = [latex]BILsin\theta[/latex]
Case I:
When current carrying conductor is placed parallelly with the direction of magnetic field, then, [latex]\theta[/latex] = 0o.
The magnitude of force becomes, Fm = BILSin0o.
[latex]\therefore[/latex] Fm = 0.
Which is called minimum force.
Case II:
When a current carrying conductor is kept perpendicularly with the direction of magnetic field, then, [latex]\theta[/latex] = 90o. The magnitude of magnetic force becomes,
Fm = BILSin90o
[latex]\therefore[/latex] Fm = BIL.
Which is called maximum force.
Torque on a rectangular coil in a uniform magnetic field: (where plane of a coil making an angle q with field):
When a certain amount of electric current is passed through the rectangular coil, which is kept inside the uniform magnetic field. The magnetic force produced on the coil generate torque which rotate the coil inside the field. The area of the plane of the coil becomes perpendicular to the field direction.
Let, ABCD be the rectangular coil having length ‘l’ and breadth ‘b’. F1 and F2 represent inward and outward force produced on a rectangular coil of sides AB and CD respectively. The normal of planes of a coil ‘n’ makes an angle [latex]\theta[/latex] with magnetic field and plane of the coil makes angle [latex]\theta^{‘}[/latex] with the direction of field. AB and CD represent current carrying conductor inside a field i.e. the force produced on a current carrying conductor inside the uniform magnetic field is given by:
[latex]\vec{F} = I\times (\vec{L}\times \vec{B})[/latex]
Magnetic force act on side AB is represented by
[latex]\vec{F_1} = I\times (\vec{L}\times \vec{B})[/latex]
and magnetic force on side CD becomes F2 = [latex]I\times (\vec{L}\times \vec{B})[/latex] .
The magnetic force produced on side BC and DA equal in magnitude and opposite in direction. So, they cancel each other. The couple of force [latex]\vec{F_1}[/latex] and [latex]\vec{F_2}[/latex] produce torque on a rectangular coil which rotates it in the direction of current.
Now, torque produced on the coil,
[latex]\tau[/latex] = Either force x perpendicular distance between 2 forces
[latex]\tau = \vec{F_1}\times DN[/latex]
[latex]\tau = I \times (\vec{L}\times \vec{B})\times bsin\theta[/latex] [∵ From [latex]\Delta AND, sin\theta = \frac{DN}{AD}, gives, bsin\theta = DN[/latex]]
[latex]\tau = I(LBsin90^o)\times bsin\theta[/latex]
[latex]\tau = IB(LB)sin\theta[/latex]
[latex]\therefore \tau = BIA.sin\theta[/latex] ……………… (i)
Eqn. (i) represent expression for torque in a single turn of coil.
For ‘N’ no. of turns in a rectangular coil, the total torque produced on it becomes [latex]\tau_N = N\tau[/latex]
Or, [latex]\tau_N = NBIAsin\theta[/latex]
[latex]\therefore \tau_N = BINAsin\theta[/latex]
[latex]\theta^{‘}[/latex] represents the angle between plane of the coil with magnetic field. Hence, from figure, we get,
[latex]\theta + \theta^{‘}[/latex] = 90o
[latex]\theta = 90^o – \theta^{‘}[/latex]
Then, above expression becomes,
[latex]\tau_N’ = BINAsin(90^o – \theta^{‘})[/latex]
= [latex]BINAcos\theta^{‘}[/latex]
Special Cases:
i) When the rectangular coil is kept parallelly with the direction of magnetic field, then, we get, [latex]\theta^{‘}[/latex] = 0o,
[latex]\tau_N’ = BINAcos0^o[/latex]
= BINA
Which is called maximum torque produced on the coil.
ii) When the rectangular coil is kept perpendicularly with the direction of magnetic field, then, we get, [latex]\theta^{‘}[/latex] = 90o,
[latex]\tau_N’ = BINAcos90^o[/latex]
= 0
Which is called minimum torque produced on a rectangular coil.
Moving Coil Galvanometer:
Moving coil galvanometer is an electrical device which is used to detect and measure the small value of electric current. It is called the basic component of ammeter and voltmeter. It is highly sensitive device.
It is based on the principle that when certain amount of electric current is passed through the rectangular coil kept inside the uniform magnetic field, it experiences a torque.
Let us consider, a rectangular coil ABCD having side length L and b which is kept inside the two cylindrical permanent magnetic poles. The normal of the plane makes an angle [latex]\theta[/latex] which is equal to 90o. The torque experienced by the rectangular coil due to the couples of two forces becomes,
[latex]\tau_N = BINAsin\theta[/latex]
For, [latex]\theta[/latex] = 90o,
[latex]\tau_N[/latex] = BINA …………… (i)
Which is called deflecting torque.
When a rectangular coil is deflected from its original position, the wire of suspension is twisted and certain restoring torque is produced.
Let, k be the restoring torque per unit twist. i.e. for angle of deflection α, the value of restoring torque becomes, [latex]\tau_N’ = k\alpha[/latex] ………………… (ii)
For an equilibrium condition, deflecting torque is equal to restoring torque, i.e.
[latex]\tau_N = \tau_N'[/latex]
BINA = kα
I = [latex]\frac{k}{BAN}\alpha[/latex]
I = Gα, where, G = [latex]\frac{k}{BAN’}[/latex], which is called Galvanometer constant.
I ∝ α.
Hence, angle of deflection is directly proportional to amount of electric current passing through the coil which can be measured by using a linear scale.
Sensitivities on Galvanometer:
Galvanometer is a highly sensitive device which contains 2 types of sensitivities:
i) Current Sensitivity:
It is defined as angle of deflection of a rectangular coil per unit electric current passing through it. It is denoted by [latex](\frac{\alpha}{I})[/latex].
We know that,
BINA = kα.
Or, [latex](\frac{\alpha}{I}) = \frac{BAN}{k}[/latex]
Or, [latex](\frac{\alpha}{I})[/latex] = current sensitivity = [latex]\frac{BAN}{k}[/latex].
Current sensitivities increases when the
i) magnetic field strength is increased.
ii) The area of the coil is increased.
iii) No. of turns used in a coil is increased.
iv) Restoring torque per unit twist is decreased.
ii) Voltage Sensitivity:
It is defined as the angle of deflecting produced on a rectangular coil per unit potential difference developed across it. It is denoted by [latex](\frac{\alpha}{V})[/latex].
We know that,
V.S. or [latex](\frac{\alpha}{V}) = (\frac{\alpha}{IR})[/latex]
Or, [latex](\frac{\alpha}{V}) = (\frac{\alpha / I}{R})[/latex]
[latex]\therefore (\frac{\alpha}{V}) = \frac{BAN}{kR}[/latex] .
Voltage sensitivity increases when the
i) Magnetic field strength is increased.
ii) The area of coil is increased.
iii) No. of turns used in a coil is increased.
iv) Restoring torque per unit twist is decreased.
v) Resistivity of coil decreases.
Hall Effect
When a magnetic field is applied to a current carrying conductor, a voltage is developed across the specimen in a perpendicular direction of flow of electric current and applied magnetic field. Such phenomenon represents hall effect.
Let us consider, a rectangular specimen through which electric current is passed in the positive x-direction which is Ix. Magnetic field is applied across z-axis where emf is produced along y-axis in a perpendicular direction of Ix and Bz which is called Hall voltage.
Before the application of magnetic field due to electric field, the electrons are drifted along negative x-axis with velocity vx. After the application of magnetic field, a certain magnetic force is experienced by the electron which is called Lorentz magnetic force. Due to Lorentz force, electrons start to deviate towards lower side of metallic plate and accumulate at the bottom. The upper side of the metallic plate contains positive charge only due to deficiency of electron between positive and negative charge, a certain electrostatic field is developed which is called hall field.
The magnitude of the Lorentz magnetic force when electrons are drifted vertically downwards is F1 = evxBz
The process of accumulating electrons in the lower side of the metallic plate is stopped when hall field balance the Lorentz magnetic force, i.e., for equilibrium condition,
FH = FL
Or, eEH = evxBz
Or, EH = [latex]\frac{ev_xB_z}{e}[/latex] …………….. (i)
According to Ohm’s law, current density is defined as the amount of electric current per unit cross sectional area.
JX = -envx ……… (ii)
Where, -ve sign indicates electronic charge and n represents total no. of free electrons per unit volume.
Dividing eqn. (i) by (ii), we get,
[latex]\frac{E_H}{J_x} = \frac{V_xB_z}{-env_x}[/latex]
EH = [latex]\frac{1}{-ne}J_xB_z[/latex]
Hence, hall field is directly proportional to current density and transverse magnetic field. Now,
[latex]\frac{E_H}{J_xB_z} = -\frac{1}{me}[/latex]
= RH, where, RH represent Hall coefficient or hall constant. It is defined as the ratio of electric field intensity per unit current density with transverse magnetic field.
Biot-Savert’s law
When electric current is passed through the conductor, it produced certain magnetic field around it. This is the basic concept of Biot-Savert’s law.
Consider a current element AB of a thin curved conductor XY shown in figure through which a constant current (I) is maintained. Let dB be the magnitude of the magnetic field [latex]\vec{dB}[/latex] at point P at radial distance ‘r’. According to Biot-Savert’s law,
i) directly proportional to magnitude of electric current. i.e.
dB ∝ I.
ii) directly proportional to length of current carrying element. i.e.
dB ∝ dl.
iii) directly proportional to Sine of angle between dl and r. i.e.
dB ∝ Sin[latex]\theta[/latex].
iv) inversely proportional to square of the distance between C to P. i.e.
dB ∝ [latex]\frac{1}{r^2}[/latex]
Combining these four relations, we get,
dB ∝ [latex]\frac{I dl sin\theta}{r^2}[/latex]
Or, dB = K [latex]\frac{I dl sin\theta}{r^2}[/latex]
Where, K = proportionality constant.
In SI system, K = [latex]\frac{\mu_o}{4\pi}[/latex]
Therefore, dB = [latex]\frac{\mu_o I dl sin\theta}{4\pi r^2}[/latex]
Which represent expression for magnitude of magnetic field strength produced at point P.
Vector or Mathematical form of dB:
Let, [latex]\vec{r}[/latex] represents the displacement of magnetic field producing point from small element. dl represents length of current carrying element in a specific direction [latex]\hat{r}[/latex] represents unit vector of r. Then,
[latex]\vec{dB} = \frac{\mu_oI(\vec{dl}\times \hat{r})}{4\pi r^2}[/latex]
The unit vector of r becomes,
[latex]\hat{r} = \frac{\vec{r}}{r}[/latex]
Or, [latex]\vec{dB} = \frac{\mu_oI(\vec{dl}\times \vec{r})}{4\pi r^3}[/latex]
Again, magnitude of magnetic field strength becomes,
dB = [latex]|\vec{dB}| = \frac{\mu_oI|\vec{dl}\times \vec{r}|}{4\pi r^3}[/latex]
Application of Biot-Savert’s Law:
1. To find the magnetic field produced on the centre of current carrying circular coil
Let us consider, a current carrying circular coil having centre O and radius ‘r’. When ‘I’ amount of electric current is passed through the current carrying coil. A certain magnetic field is produced at point O. ACB represents small element of current carrying coil which produced certain magnetic field at point O at a distance r from its centre which is equivalent to radius.
According to Biot-Savert’s law, the magnetic field strength produced by small element is:
dB = [latex]\frac{\mu_o Idlsin\theta}{4\pi r^2}[/latex]
In case of circular coil, the angle between dl and r is 90o, i.e.
[latex]\theta[/latex] = 90o.
[latex]\therefore dB = \frac{\mu_oIdlsin90^o}{4\pi r^2}[/latex]
dB = [latex]\frac{\mu_oIdI}{4\pi r^2}[/latex]
To obtain total magnetic field strength produced by whole part of the current carrying circular coil, we should integrate the above expression from limit 0 to 2πr. i.e.
[latex]\int_0^{2\pi r}{dB} = \int_0^{2\pi r}{\frac{\mu_o Idl}{4\pi r^2}}[/latex]
Or, B = [latex]\frac{\mu_o I}{4\pi r^2}\int_0^{2\pi r}{dl}[/latex]
Or, B = [latex]\frac{\mu_o I}{2r}\times 2\pi r[/latex]
[latex]\therefore B = \frac{\mu_oI}{2r}[/latex] …………….. (i)
Hence, eqn. (i) represents magnetic field strength produced at the centre by single turn of coil.
For ‘N’ no. of turns, the expression of magnetic field strength becomes,
B = [latex]\frac{\mu_oNI}{2r}[/latex]
2. Magnetic field produced on the axis of current carrying coil
Let us consider, a current carrying circular coil having centre O and radius ‘a’. When ‘I’ amount of electric current is passed through the circular coil. It produced certain through the circular coil. It produced certain magnetic field around it at a point ‘P’ which is ‘x’ distance far from the centre.
According to Biot-Savert law, the magnitude of small magnetic field strength produced at a point is dB i.e.
dB = [latex]\frac{\mu_o IdL sin\theta}{4\pi r^2}[/latex]
For the negligible radius of the circular coil, the angle between dl and r becomes [latex]\frac{\pi}{2}[/latex]. i.e.
dB = [latex]\frac{\mu_o IdL sin\pi /2}{4\pi r^2}[/latex]
[latex]\therefore dB = \frac{\mu_o IdL}{4\pi r^2}[/latex].
The magnetic field strength act along PL where, dB makes an angle 90o with PC. The small magnetic field strength dB resolves into 2 components.
Vertical component represented by dBCos[latex]\theta[/latex].
Horizontal component represented by dBSin[latex]\theta[/latex].
As the current carrying circular coil is symmetrical about its axis, every element of length [latex]dl[/latex] has equal and opposite element. The magnetic field strength produced by A’B’ is dB’ which has 2 components:
dB’Cos[latex]\theta[/latex] act vertically and
dB’Sin[latex]\theta[/latex] act horizontally.
For the same magnitude of dB and dB’, there is no contribution of vertical component as they cancel each other but horizontal components dBsin remains all along the axis of the coil.
Hence, to obtain total magnetic field produced by the circle at point P, we should integrate the horizontal component from the limit 0 – 2πa. i.e.
B = [latex]\int_o^{2\pi a}dBsin\theta[/latex]
Or, B = [latex]\int_o^{2\pi a} \frac{\mu_o I dL}{4\pi r^2}sin\theta[/latex]
= [latex]\frac{\mu_o I sin\theta}{4\pi r^2}\int_0^{2\pi a}dl[/latex]
= [latex]\frac{\mu_oIsin\theta}{4\pi r^2}\times 2\pi a[/latex]
= [latex]\frac{\mu_o I sin\theta . a}{2r^2}[/latex]
= [latex]\frac{\mu_o I asin\theta}{2r^2}[/latex]
From [latex]\Delta[/latex]POC,
[latex]sin\theta = \frac{a}{r}[/latex]
Then, B = [latex]\frac{\mu_o Ia^2}{2r^3}[/latex]
[latex]\therefore B = \frac{\mu_o I a^2}{2(a^2 + x^2)^{\frac{3}{2}}}[/latex] …………….. (i)
Hence, eqn. (i) represents the expression of magnetic field produced at point P by single turn of circular coil. For ‘N’ no. of turns, the above expression can be written as,
B = [latex]\frac{\mu_oINa^2}{2(a^2 + x^2)^{\frac{3}{2}}}[/latex]
Special Cases:
(I) when the magnetic field producing point lies at a centre of circular coil. i.e.
x = 0
[latex]\therefore B = \frac{\mu_o I Na^2}{2(a^2 + 0)^{\frac{3}{2}}}[/latex]
= [latex]\frac{\mu_oINa^2}{2a^3}[/latex]
= [latex]\frac{\mu_o IN}{2a}[/latex], which is the same expression of the magnetic field produced at the centre of circular coil.
(II) When magnetic field producing point lies so far from the radius of the coil. i.e.
x ≫ a.
[latex]\therefore B = \frac{\mu_o INa^2}{2(x^2)^{\frac{3}{2}}}[/latex]
Or, B = [latex]\frac{\mu_o INa^2}{2x^3}[/latex]
3) Magnetic field produced by infinitely long current carrying conductor:
Let us consider, a current carrying straight wire AB have a finite length. When I amount of electric current is passed, it produced magnetic field strength near a wire at a point P which is ‘a’ distance far from the wire. ‘dl’ represents small current carrying element which produced magnetic field at a point P which lies at a distance ‘r’ from it.
[latex]\phi[/latex] represent the angle between dl and r.
According to Biot-Savert law, the magnetic field strength produced by small element is given by:
dB = [latex]\frac{\mu_o Idl sin\phi}{4\pi r^2}[/latex] …………… (i)
Then, from triangle [latex]\Delta[/latex]POC,
[latex]sin\theta = \frac{a}{r} = cos\theta[/latex]
Or, [latex]cos\theta = \frac{a}{r}[/latex]
Or, r = [latex]\frac{a}{cos\theta}[/latex]
Or, r = [latex]asec\theta[/latex]
Similarly, [latex]tan\theta = \frac{I}{a}[/latex]
Or, [latex]I = atan\theta[/latex]
Differentiating above expression with respect to [latex]\theta[/latex],
[latex]\frac{dl}{d\theta} = asec^2\theta[/latex]
[latex]\therefore dI = Asec^2 d\theta[/latex]
Then, above expression (i) becomes,
dB = [latex]\frac{\mu_o Ia sec^2\theta d\theta cos\theta}{4\pi a^2 sec^2\theta}[/latex]
= [latex]\frac{\mu_o I}{4\pi a}cos\theta d\theta[/latex]
To obtain total magnetic field produced by finitely long current carrying conductor at point P, we should integrate the above expression from limiting value [latex]-\theta_1[/latex] to [latex]\theta_2[/latex] (Sign convention determined by direction)
B = [latex]\int_{-\theta_1}^{\theta_2}dB[/latex]
Or, B = [latex]\int_{-\theta_1}^{\theta_2} \frac{\mu_o I}{4\pi a}cos\theta d\theta[/latex]
Or, B = [latex]\frac{\mu_o I}{4\pi a}\int_{-\theta_1}^{\theta_2}cos\theta d\theta[/latex]
Or, B = [latex]\frac{\mu_o I}{4\pi a}[sin\theta]_{-\theta_1}^{\theta_2}[/latex]
Or, B = [latex]\frac{\mu_o I}{4\pi a}[sin\theta_2 – (sin(-\theta_1))][/latex]
[latex]\therefore B = \frac{\mu_o I}{4\pi a}[sin\theta_1 + sin\theta_2][/latex] ………………… (ii)
Hence, eqn. (ii) represents the total magnetic field strength produced by finitely long wire AB.
For infinitely long straight wire, [latex]\theta_1[/latex] and [latex]\theta_2[/latex] is replaced by [latex]\frac{\pi}{2}[/latex] or 90o.
Then, B = [latex]\frac{\mu_o I}{4\pi a}[sin90^o + sin90^o][/latex]
= [latex]\frac{\mu_o I}{4\pi a}[1 + 1][/latex]
= [latex]\frac{\mu_o I.2}{4\pi a}[/latex]
[latex]\therefore B = \frac{\mu_o I}{2\pi a}[/latex] …………….. (iii)
Hence, eqn. (iii) represents the expression for magnetic field strength produced by infinitely long straight current carrying conductor.
Ampere Circuital Law
The line integral of magnetic field around any closed path or circuit in vacuum is equal to [latex]\mu_o[/latex] (absolute permeability of free space) times the total current enclosed by this path.
Mathematically, [latex]\oint \vec{B}.\vec{dl} = \mu_o I[/latex], where, B is magnetic field intensity due to current [latex]I[/latex] enclosed by a close path.
Applications of Ampere’s Circuital Law
1. Magnetic field due to a straight current carrying conductor
Let us consider an infinitely long straight conductor carrying current [latex]I[/latex] as shown in figure. Now, we want to find magnetic field at a point P at perpendicular distance ‘a’ from the conductor. A circle of radius ‘a’ with centre at the conductor is drawn. By symmetry magnetic field at all points of the conductor must be same and is tangential to the circle of that point, [latex]\vec{B}[/latex] and [latex]\vec{dl}[/latex] are in same direction so [latex]\theta = 0^o[/latex].
From Ampere’s circuital law,
[latex]\oint \vec{B}.\vec{dl} = \mu_o I[/latex]
Or, [latex]\oint Bdl cos0^o = \mu_o I[/latex]
Or, [latex]B\oint \vec{dl} = \mu_o I[/latex]
Since, [latex]\oint \vec{dl}[/latex] = circumference = [latex]2\pi a[/latex], we have,
[latex]B(2\pi a) = \mu_o I[/latex]
[latex]\therefore B = \frac{\mu_o I}{2\pi a}[/latex]
This is the required expression for magnetic field due to an infinitely long current carrying conductor.
2. Magnetic field due to a current carrying solenoid
Let us consider a long solenoid carrying current ‘[latex]I[/latex]’ and possessing ‘n’ number of turns per unit length. Magnetic field outside the solenoid is neglected since very weak but magnetic field inside the solenoid is uniform, strong and directed along the axis of the solenoid.
Now, let us take a rectangular path abcd as shown in figure. It encloses ‘nh’ number of turns of the solenoid.
Here,
Total current enclosed by the closed loop = [latex]nhI[/latex] …………… (i)
Again, line integral of B around the close path ‘abcd’ is given by,
[latex]\oint \vec{B}.\vec{dl} = \int_{a}^{b} \vec{B}.\vec{dl} + \int_{b}^{c} \vec{B}.\vec{dl} + \int_{c}^{d} \vec{B}.\vec{dl}+\int_{d}^{a}\vec{B}.\vec{dl}[/latex] ……………. (ii)
where, [latex]\int_{a}^{b} \vec{B}.\vec{dl} = \int_{a}^{b} Bdlcos0^o = B\int_{a}^{b}dl = Bh[/latex]
[latex]\int_{b}^{c} = \int_{b}^{c} Bdl cos90^o = 0[/latex]
[latex]\int_{c}^{d} \vec{B}.\vec{dl} = \int_{c}^{d} Bdl = 0[/latex]
[[latex]\because[/latex]B = 0 outside solenoid]
[latex]\int_{d}^{a}\vec{B}.\vec{dl} = \int_{d}^{a}Bdlcos90^o = 0[/latex]
Substituting these values in equation (ii), we get, [latex]\oint \vec{B}.\vec{dl} = Bh[/latex] ……………. (iii)
Again, from Ampere’s circuital law, we get,
[latex]\oint \vec{B}.\vec{dl} = \mu_o \times[/latex] (Enclosed current) …………. (iv)
From equation (i), (iii) and (iv),
[latex]Bh = \mu_o nhI[/latex]
[latex]\therefore B = \mu_o nI[/latex]
This is the required expression for magnetic field due to a current carrying solenoid.
3. Magnetic field due to a current carrying toroid
Let us consider, a toroidal solenoid of radius ‘a’ possessing ‘n’ number of turns per unit length and carrying current [latex]I[/latex]. A point P is considered (inside the toroid windings) at distance r from centre ‘O’ of the toroid as shown in figure. A circle of radius ‘a is drawn which passes through point P.
According to Ampere’s circuital law, we have,
[latex]\oint \vec{B}.\vec{dl} = \mu_o I_o[/latex] ……………. (i) where, [latex]I_o[/latex] is the total current enclosed by the circle passing through point P. So, [latex]I_o = NI[/latex], where N is total number of turns of the toroidal solenoid.
Now, [latex]\oint \vec{B}.\vec{dl} = \oint Bdl cos\theta[/latex]
Since, [latex]cos\theta = 1[/latex]
[latex]\therefore Bdl = B\oint dI = B(2\pi a)[/latex]
Hence, equation (i) can be written as,
[latex]B(2\pi a) = \mu_o NI[/latex]
Or, [latex]B(2\pi a) = \mu_o (2\pi an)I[/latex]
Or, B = [latex]\mu_o nI[/latex]
This is the required expression for magnetic field due to a current carrying toroid.
Force between two parallel current carrying conductors when currents pass in same direction
Let us consider, two parallel conductors X and Y at distance ‘r’ apart, carrying current [latex]I_1[/latex] and [latex]I_2[/latex] in the same direction, take two points P and Q on conductors Y and X respectively as shown in figure.
Now, magnetic flux density at point P due to current [latex]I_1[/latex] in X is given as [latex]B_X = \frac{\mu_o I_1}{2\pi r}[/latex] ……… (i), its direction is inward and perpendicular to the plane of the paper.
Now the force acting on conductor Y due to magnetic field [latex]B_X[/latex] is
[latex]F_Y = B_XI_2Isin\theta = B_XI_2[/latex] ………… (ii) [[latex]\because \theta = 90^o[/latex] and taking [latex]l = 1[/latex]]
Then, force per unit length of the conductor Y due to the magnetic field produced by X is given as
[latex]F_Y = \frac{\mu_o I_1I_2}{2\pi r}[/latex] ………….. (iii) [from equations (i) and (ii)]
This force is directed towards the conductor X in the plane of paper according to Fleming’s left-hand rule.
Similarly, magnetic flux density at point Q due to current [latex]I_2[/latex] in Y is given as [latex]B_Y = \frac{\mu_oI_2}{2\pi r}[/latex] ………. (iv), its direction is outward and perpendicular to the plane of the paper.
Now the force acting on conductor X due to magnetic field [latex]B_Y[/latex] is
[latex]F_X = B_YI_1Isin\theta = B_YI_1[/latex] ………… (v) [[latex]\theta = 90^o[/latex] and taking [latex]l = 1[/latex]]
Then, force per unit length of the conductor X due to the magnetic field produced by Y will be
[latex]F_X = \frac{\mu_o I_2 I_1}{2\pi r}[/latex] ………….. (vi) [from equations (iv) and (v)]
This force is directed towards conductor X in the plane of the paper and it is attractive in nature.
When, currents pass in opposite direction, then, [latex]F_X = – F_Y[/latex]. It is repulsive in nature.
Definition of one ampere
Force per unit length on current carrying parallel conductors due to the magnetic field generated by each other is given by,
[latex]F = \frac{\mu_o I_1 I_2}{2\pi r}[/latex]
When [latex]I_1 = I_2 = 1[/latex] ampere, r = 1 m in vacuum, we get,
[latex]F = \frac{4\pi \times 10^{-7}\times 1 \times 1}{2\pi \times 1}[/latex]
= [latex]2\pi \times 10^{-7}\ N[/latex]
Hence, one ampere current is defined as the current which when passed through two infinitely long parallel conductors separated by one meter in vacuum given rise to force of [latex]2\times 10^{-7}[/latex] N per meter length on each other.
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