Contents
- 1 What is the Working Principle of Barlow’s Wheel? What is the Torque on Wire in a Uniform Magnetic Field?
- 1.1 Force on a Current Carrying Conductor in a Magnetic Field
- 1.2 Special cases:
- 1.3 Discussions:
- 1.4 Torque on a Current Loop in a Uniform Magnetic Field
- 1.5 Numerical Examples
- 1.6 Action of Current on Current
- 1.7 Oblique currents:
- 1.8 Experimental Demonstration : Roget’s Vibrating Spiral
- 1.9 Numerical Examples
Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.
What is the Working Principle of Barlow’s Wheel? What is the Torque on Wire in a Uniform Magnetic Field?
Oersted’s experiment deals with the influence of a current car-rying conductor on a magnet. The magnet, in reaction, also exerts equal but opposite force on the current carrying con-ductor. So, it is possible to move a current carrying wire in a magnetic field. This kind of motion is very important for practi-cal purposes. This phenomenon is the subject matter of electro-dynamics.
Force on a Current Carrying Conductor in a Magnetic Field
The Lorentz force acting on a charge q moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is,
\(\vec{F}\) = q\(\vec{v}\) × \(\vec{B}\) (see section 1.6)
When current flows in a circuit, the drift motion of free electric charges through the circuit is considered as the cause of that current. Naturally the magnetic force acting on those free charges in a magnetic field acts as the force on the current carrying conductor.
Let the current through a circuit = I; an element of length dt of that circuit is considered [Fig.].
Magnetic field at the position of this element of the circuit = \(\vec{B}\). If dq amount of charge crosses dl length in time dt, current in the circuit, I = \(\frac{d q}{d t}\);
velocity of the charge, \(\vec{v}\) = \(\frac{d \vec{l}}{d t}\).
So, the magnetic force acting on the small part \(d \vec{l}\),
\(d \vec{F}\) = \(d q \vec{v} \times \vec{B}\) = \(d q \cdot \frac{d \vec{l}}{d t} \times \vec{B}\) = \(\) …. (1)
If the angle between \(d \vec{l}\) and \(\vec{B}\) be θ, then
dF = |\(d \vec{F}\)| = IdlBsinθ = BIdlsinθ …….. (2)
So, the magnetic force acting on the whole circuit or on a finite part of it,
F = \(\int d F\) = \(\int B I d l \sin \theta\) …… (3)
Generally, for different elements of the circuit, both B and θ may change. Hence, in absence of some special symmetry the integral of equation (3) becomes very complicated.
Special cases:
i) The current carrying wire is at right angles with a uni-form magnetic field: For any small part of the wire, θ = 90° or sin θ = 1 [Fig.(a)]. If the magnetic field is uniform, B = constant. If a current carrying wire of length l is placed in this uniform magnetic field, according to equation (3) the force acting on the wire will be,
F = BIl ……. (4)
ii) The current carrying wire is placed parallel to the mag-netic field: For any small part of the wire, θ = 0° or sin θ = 0 [Fig.(b)],
So, according to equation (3), the force acting on the wire is,
F = 0 ……. (5)
Note that, the units of l ox dl, I, B and F are m, A, Wb ᐧ m-2 and N, respectively.
Fleming’s left hand rule: If a long straight current car-rying conductor is kept at right angles with a uniform magnetic field, i.e., if for each element of the wire, the angle between \(\overrightarrow{d l}\) and B is θ = 90°, then according to the rule of cross-product we can say from equation (1) that the three vectors \(\overrightarrow{d l}, \vec{B}\) and \(d \vec{F}\) will be mutually perpendicular. Fleming’s left hand rule is a handy tool for determination of the direction of the magnetic force acting on a current carrying conductor, perpendicular to the magnetic field.
Statement: If the forefinger, middle finger and the thumb of the left hand are stretched mutually perpendicular to each other, such that the forefinger points the direction of the mag-netic field and the middle finger points the direction of current,
then the thumb points the direction of force experienced by the conductor [Fig.].
This left hand rule is also known as motor rule.
Barlow’s wheel: We can demonstrate action of magnet on electric current with the help of this experimental arrangement.
Description: B is the Barlow’s wheel made of thin copper plate having a number of sharp teeth [Fig. 1.64]. The wheel B is kept vertical such that its one sharp tooth remains in contact with mercury kept in a container placed in between two poles of a strong horse-shoe magnet (NS). Now the wheel and the mer-cury inside the container (M) are connected to a source of cur-rent E.
Working principle: The magnetic field acts from Af-pole to S-pole of the magnet parallel to the horizontal platform. Since the tooth of the wheel remains in contact with mercury, the circuit is closed as soon as the battery is switched on. As a result, current flows in a downward direction.
From Fleming’s left hand rule we see that, a magnetic force acts on the current towards right. Due to this force, the tooth will be deflected towards right. As a result, the circuit breaks. Hence magnetic force will no longer act on the wheel. But due to inertia of motion, the next tooth comes in contact with mercury and in a similar way it will be deflected towards right. Hence, a contin-uous anticlockwise motion will be observed in the Barlow’s wheel.
Discussions:
- To get effective result, the wheel should be very light. Moreover, the magnetic field and the electric current should be very strong.
- If the direction of either the magnetic field or the electric current be reversed, the rotation of the wheel will also be reversed; but if both of them are reversed then the rotation of wheel will be in the same direction.
- The machine which converts electrical energy into rotational energy (usually mechanical energy) is called a motor. So, Barlow’s wheel is a motor, although it has no practical use.
Torque on a Current Loop in a Uniform Magnetic Field
PQRS is a rectangular conductor [Fig.]. Its length, PQ = RS = l and breadth, QR= PS = b.
So, the area of the rectangular face of the conductor, A = lb.
This rectangular conductor is placed in a magnetic field \(\vec{B}\) in such a way that
1. PQ mid RS are perpendicular to the mag-netic field and
2. the surface PQRS is parallel to the magnetic field.
If a current I is sent through the conductor,
1. no force acts on the arms QR and PS as they are parallel to the magnetic field,
2. magnitude of the force (\(\vec{F}\)) acting on each of the arms PQ and RS, F = BIl.
Applying Fleming’s left hand rule, we see that these two equal forces are downward and upward, respectively and hence they constitute a couple.
Since, the perpendicular distance between these two forces is QR = PS = b, the moment of the couple, i.e., torque acting on the coil is,
\(\tau\) = BIl ᐧ b = BIA ………. (1)
and due to this torque the coil starts to rotate.
Instead of a single turn, if the conducting coil has N turns, torque acting on it,
\(\tau\) = BNIA ………… (2)
Equations (1) and (2) show that, the torque r depends only on area A of the coil. This means that the shape of the coil is not important. Instead of a rectangular coil, a coil of any other shape of equal area may be used to obtain the same torque.
Vector representation of torque: We know that, any plane surface can be treated as a vector, such that the magnitude of the vector is equal to the magnitude of its area and its direction is along the normal to that plane. So, the quantity A in equation (1) can be expressed as a vector \(\vec{A}\). Fig. shows the direction of A at an arbitrary position of the rotating loop under the influence of torque. If the angle between \(\vec{A}\) and the magnetic field \(\vec{A}\) be θ, from equation (1) we get,
\(\tau\) = BIA sin θ …… (3)
The vector representation of the above equation is,
\(\tau\) = \(I \vec{A} \times \vec{B}\) …… (4)
So, the torque acting on a conducting coil having N turns will be
\(\tau\) = \(N I \vec{A} \times \vec{B}\) …….. (5)
Numerical Examples
Example 1.
2A current is flowing through a circular coil of radius 10 cm, made of insulated wire and having 100 turns.
(i) If the circular plane of the conductor is kept at right angles to the direction of a magnetic field of 0.2 Wb ᐧ m-2, determine the force acting on the coil.
(ii) If the conductor is placed parallel to the magnetic field, determine the torque acting on it.
Solution:
Here, N = 100, r = 10 cm = 0.1 m,
B = 0.2 Wb ᐧ m-2
∴ Circumference of the circular loop,
L = 2πr = 2 × π × 0.1 = 0.628 m
Area of the circular loop,
A = πr2 = π × (0.1)2 = 0.0314 m2
i) Whole circumference of the conductor is normal to the magnetic field; hence magnetic force
= NILB = 100 × 2 × 0.628 × 0.2 = 25.12 N
[It should be noted that, in this situation if the circular plane be taken as a vector, its direction will be perpendicular to that plane, i.e., parallel to the magnetic field. As a result, no torque will act on the conductor.]
ii) If the circular plane is taken as a vector, its direction will be perpendicular to the plane, i.e., normal to the magnetic field. As a result, torque acting on the conductor
= NIAB = 100 × 2 × 0.0314 × 0.2 = 1.256 N ᐧ m
Example 2.
The radius of a circular coil having 100 turns is 5 cm and a current of 0.5 A is flowing through this coil. If it is placed in a uniform magnetic field of strength 0.001 T, then what torque wifi act on the coil, when the plane of the coil is
(i) parallel to the magnetic field,
(ii) inclined at 30° with the magnetic field,
(iii) per-pendicular to the magnetic field?
Solution:
Number of turns = 100
Current, I = 0.5 A
Radius of the circular coil, r = 5 cm = 5 × 10-2 m
∴ Area of the coil, A = πr2 = 3.14 × (5 × 10-2)2m2
Magnetic field intensity, B = 0.001 T
Now torque acting on the current carrying coil due to magnetic field, \(\tau\) = nBIAsinθ
i) When the coil is parallel to the magnetic field, then θ = 90°.
So, \(\tau\) = nBIA sin 90°
= 100 × 0.001 × 0.5 × 3.14 × 25 × 10-4
= 3.93 × 10-4 N ᐧ m
ii) When the coil makes an angle 30° with the field, then
θ = 60°
So, \(\tau\) = nBIA sin 60°
= 100 × 0.001 × 0.5 × 3.14 × 25 × 10-4 × \(\frac{\sqrt{3}}{2}\)
= 3.4 × 10-4 N ᐧ m
iii) When the coil is perpendicular to the field, θ = 0°.
∴ \(\tau\) = 0
Example 3.
On a smooth plane inclined at 30° with the horizontal, a thin current carrying metallic rod is placed parallel to the horizontal ground. The plane is in a uniform magnetic field of 0.15 T along the vertical direction. For what value of current can the rod remain station-ary? The mass per unit length of the rod is 0.30 kg/m.
Solution:
Along the inclined plane, component of the magnetic force acting on the conductor and the component of weight of the conductor will bring equilibrium. So,
BIlcosθ = mgsinθ ……… (1)
Here, force on the conductor Bil acts horizontally towards right.
Action of Current on Current
Like parallel currents: Let two straight parallel conductors PQ and RS be kept horizontally and I1 and I2 be the currents flowing through them in the same direction [Fig.]
According to corkscrew rule, for current I1 the magnetic field B at any point O on the wire RS will act downwards. If Flemings left hand rule is applied at the point O on wire RS, it is seen that the wire RS will experience a force F towards the wire PQ.
As reaction, the wire PQ also will experience the same force F towards the wire RS. So, we can conclude that two like parallel currents auract each other.
Unlike parallel currents: Suppose I1 and I2 currents are flowing through the wires PQ and RS, respectively in mutually opposite directions. Applying Fleming’s left hand rule similarly, it is found that the wire RS experiences a force F away from the wire PQ. The wire PQ also experiences an equal but opposite force that acts on it away from the wire RS [Fig.]. So, we can conclude that two unlike parallel currents repel each other.
Magnitude of torce of attraction or repuision: Let r be the perpendicular distance between two long, straight parallel conductors kept in vacuum or air [Fig.].
Due to current I the magnetic field at a distance r will be,
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{r}\)
From the relation F = BIl we get, force acting on unit length of the wire placed at r, carrying current I2 is
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}\) [Putting B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{r}\) and I = I2]
Here, the unit of r is m, I1 and I2 is A, F is N ᐧ m-1 and µ0 = 4π × 10-7 H ᐧ m-1.
This equation gives the definition of international ampere. If I1 = I2 = 1A and r = 1 m,
F = \(\frac{\mu_0}{4 \pi} \times \frac{2 \times 1 \times 1}{1}\) = \(\frac{4 \pi \times 10^{-7}}{4 \pi}\) × 2 = 2 × 10-7 N
1 ampere: Two long straight parallel conducting wires, having negligible cross sections and carrying equal currents are kept 1 m apart from each other. The steady direct current for which each wire experiences a force of 2 × 10-7 N per unit length, is called 1 ampere.
Expressions in Gaussian system: Substituting I1 → i1, I2 → i2 and μ0 → 4π in equation (1) we get,
F = \(\frac{2 i_1 i_2}{r}\)
Here the unit of r is cm, F is dyn and i1 and i2 is emu of current.
Oblique currents:
i) If currents through two oblique straight conductors lying in the same plane converge to or diverge from their point of intersection, they attract each other [Fig.].
ii) If currents through two oblique straight conductors lying in the same plane are such that, one of them is directed towards and the other directed away from their point of intersection, they repel each other [Fig.].
Experimental Demonstration : Roget’s Vibrating Spiral
Description: A long elastic spring made of insulted copper wire is suspended from a rigid support [Fig]. A small spherical copper bob, attached at the lower end of the spring, just touches the mercury kept in a vessel. The fulcrum at the upper end of the spring and mercury in the vessel are joined to a battery.
Working principle: When the battery is switched on, a unidirectional parallel current flows through the turns of the spring and hence the turns attract each other, resulting in con-traction of the spring. Naturally, the copper bob remains no lon-ger in contact with mercury and hence the circuit is cut off. As soon as the bob leaves the surface of the mercury, the attraction between the turns of the spring no longer exists. Hence, the spring elongates due to its own weight, the copper sphere touches the mercury again and closes the circuit. In this way, alternate compression and elongation of the spring goes on and hence the spring vibrates continuously.
This experiment proves that a number of unidirectional parallel currents attract each other.
Numerical Examples
Example 1.
Two very long conducting wires are kept at a distance 4 cm from each other in vacuum. Currents flowing through the wires are 25A and 5A, respectively. Find the length of each conductor, which experiences a force of 125 dynes ?
Solution:
Here, I1 = 25 A, I2 = 5 A, r = 4 cm = 0.04 m
So, the force acting per unit length of the wire,
F = F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}\) = \(\frac{4 \pi \times 10^{-7}}{4 \pi} \cdot \frac{2 \times 25 \times 5}{0.04}\)
= 625 × 10-6 N ᐧ m-1
∴ For 125 dyn or 125 × 10-5 N force, the effective length of each wire,
l = \(\frac{125 \times 10^{-5}}{625 \times 10^{-6}}\) = 2 m
Example 2.
Two long straight parallel conducting wires, kept 0.5 m apart, carry 1A and 3 A currents, respectively [Fig.].
(i)What is the force acting per unit length of the two wires?
(ii) At what position in the plane of the wires, the resultant magnetic field will be zero?
Solution:
i) Force acting per unit length,
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}\) = 10-7 × \(\frac{2 \times 1 \times 3}{0.5}\)
= 1.2 × 10-6 N ᐧ m-1
ii) According to corkscrew rule, the magnetic fields produced between two wires are mutually opposite in direction. Let any point P in this region be at a distance x from the first wire.
∴ Magnetic field at the point P due to the first wire
= \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{x}\)
and magnetic field at the point P due to the second wire
= \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_2}{(0.5-x)}\)
If the resultant magnetic field at the point P be zero, then
\(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{x}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_2}{(0.5-x)}\) or, \(\frac{0.5-x}{x}\) = \(\frac{I_2}{I_1}\) = \(\frac{3}{1}\)
or, 3x = 0.5 – x or, 4x = 0.5
or, x = \(\frac{0.5}{4}\) = 0.125 m
So, at any point on the plane between the wires, 0.125 m away from the first wire, the resultant magnetic field will be zero.
Example 3.
Two long parallel conductors, kept at a distance d, carry currents I1 and I2, respectively. The mutual force acting between them is F. Now, the current in one is doubled and its direction is also reversed. If the distance of separation between them is made 3d, what will be the force acting between the two conductors? [AIEEE ’04]
Solution:
For unit length of the conductors,
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{d}\)
If the current in the first conductor is doubled and its direction is reversed, it will be -2 I1.
∴ In the second case
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2\left(-2 I_1\right) I_2}{3 d}\) = \(-\frac{2}{3} \frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{d}\) = \(-\frac{2}{3} F\)
Example 4.
A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another par-allel long wire CD, which is fixed in a horizontal plane and carries a steady current of 30 A as shown in figure. Show that when AB is slightly depressed, it executes simple harmonic motion. Find its period of oscillations.
Solution:
Force per unit length on wire AB due to the magnetic field produced by wire CD
= \(\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r}\)
where I1 = 20 A, I2 = 30 A and distance between AB and CD, r = 0.01m.
So force on the whole of wire AB due to wire CD,
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}\) × L [where L = length of wire AB ]
Since wire AB is in equilibrium, the force F must balance its weight acting downwards. This is possible only if the two wires repel each other and hence I1 and I2 are flowing in opposite directions.
Accordingly, F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}\) × L = Mg …….. (1)
[where M =mass of wire AB ]
If we consider that the wire is depressed by a small distance d so that the height of AB over CD becomes (r – d), then the force on wire AB increases to,
Now on releasing the wire AB, it moves under the effect of restoring force f.
Acceleration of the wire AB, a = \(\frac{f}{M}\) = \(\frac{\mu_0 I_1 I_2 L}{2 \pi r^2 M} d\)
Since, except d all other physical quantities are constants a ∝ d.
Hence the motion of the wire is simple harmonic. So time period,
So time period,