Force Between Two Parallel Current , The Ampere

Force Between Two Parallel Current,  The Ampere

We have learnt that there exists a magnetic field due to a conductor carrying a current which obeys the Biot-Savart law. Further, we have learnt that an external magnetic field will exert a force on a current-carrying conductor. This follows from the Lorentz force formula. Thus, it is logical to expect that two current-carrying conductors placed near each other will exert (magnetic) forces on each other. In the period 1820-25, Ampere studied the nature of this magnetic force and its dependence on the magnitude of the current, on the shape and size of the conductors, as well as, the distances between the conductors. In this section, we shall take the simple example of two parallel current- carrying conductors, which will perhaps help us to appreciate Ampere’s painstaking work.

Figure 4.20 shows two long parallel conductors a and b separated by a distance d and carrying (parallel) currents I_a and I_b, respectively. The conductor ‘a’ produces, the same magnetic field Ba at all points along the conductor ‘b’. The right-hand rule tells us that the are placed horizontally). Its magnitude is given by Eq. [4.19(a)] or from Ampere’s circuital law,

FIGURE 4.20 Two long straight parallel conductors carrying steady currents I_a and I_b and separated by a distance d. Ba is the magnetic field set up by conductor ‘a’ at conductor ‘b’.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqamaaBa % aaleaacaWGHbaabeaakiabg2da9maalaaabaGaeqiVd02aaSbaaSqa % aiaaicdaaeqaaOGaamysamaaBaaaleaacaWGHbaabeaaaOqaaiaaik % dacqaHapaCcaWGKbaaaaaa!40E1! {B_a} = \frac{{{\mu _0}{I_a}}}{{2\pi d}}\)

The conductor ‘b’ carrying a current I_b will experience a sideways force due to the field B_a. The direction of this force is towards the conductor ‘a’ (Verify this). We label this force as F_ba, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by Eq. (4.4),

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGgb % WaaSbaaSqaaiaadkgacaWGHbaabeaakiabg2da9iaadMeadaWgaaWc % baGaamOyaaqabaGccaWGmbGaamOqamaaBaaaleaacaWGHbaabeaaaO % qaaiabg2da9maalaaabaGaeqiVd02aaSbaaSqaaiaaicdaaeqaaOGa % amysamaaBaaaleaacaWGHbaabeaakiaadMeadaWgaaWcbaGaamOyaa % qabaaakeaacaaIYaGaeqiWdaNaamizaaaacaWGmbGaaGPaVlaaykW7 % caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua % aGPaVlaaykW7caaMc8+aaeWaaeaacaaI0aGaaiOlaiaaikdacaaIZa % aacaGLOaGaayzkaaaaaaa!6F0D! \begin{array}{l} {F_{ba}} = {I_b}L{B_a}\\ = \frac{{{\mu _0}{I_a}{I_b}}}{{2\pi d}}L\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {4.23} \right) \end{array}\)

It is of course possible to compute the force on ‘a’ due to ‘b’. From considerations similar to above we can find the force F_ab, on a segment of length L of ‘a’ due to the current in ‘b’. It is equal in magnitude to F_ba, and directed towards ‘b’. Thus,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa % aaleaacaWGIbGaamyyaaqabaGccqGH9aqpcqGHsislcaWGgbWaaSba % aSqaaiaadggacaWGIbaabeaaaaa!3D7B! {F_{ba}} = - {F_{ab}}\)         (4.24)

Note that this is consistent with Newton’s third Law. Thus, at least for parallel conductors and steady currents, we have shown that the Biot-Savart law and the Lorentz force yield results in accordance with Newton’s third Law*.

We have seen from above that currents flowing in the same direction attract each other. One can show that oppositely directed currents repel each other. Thus,  Parallel currents attract, and antiparallel currents repel.

This rule is the opposite of what we find in electrostatics. Like (same sign) charges repel each other, but like (parallel) currents attract each other.

Let f_ba represent the magnitude of the force Fba per unit length. Then, from Eq. (4.23),

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaaBa % aaleaacaWGIbGaamyyaaqabaGccqGH9aqpdaWcaaqaaiabeY7aTnaa % BaaaleaacaaIWaaabeaakiaadMeadaWgaaWcbaGaamyyaaqabaGcca % WGjbWaaSbaaSqaaiaadkgaaeqaaaGcbaGaaGOmaiabec8aWjabec8a % Wjaadsgaaaaaaa!4594! {f_{ba}} = \frac{{{\mu _0}{I_a}{I_b}}}{{2\pi \pi d}}\)       (4.25)

The above expression is used to define the ampere (A), which is one of the seven SI base units.

The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 × 10^–7 newtons per metre of length.

This definition of the ampere was adopted in 1946. It is a theoretical definition. In practice, one must eliminate the effect of the earth’s magnetic field and substitute very long wires by multiturn coils of appropriate geometries. An instrument called the current balance is used to measure this mechanical force.

The SI unit of charge, namely, the coulomb, can now be defined in terms of the ampere.

When a steady current of 1A is set up in a conductor, the quantity of charge that flows through its cross-section in 1s is one coulomb (1C).

Example 4.10

The horizontal component of the earth’s magnetic field at a certain place is 3.0 ×10^–5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north?

Solution

F = I l × B

F = IlB sin\(\theta\)

The force per unit length is

f = F/l = I B sin\(\theta\)

(a) When the current is flowing from east to west,

\(\theta\) = 90°

Hence,

f = I B

= 1 × 3 × 10^–5 = 3 × 10^–5 N m^–1

This is larger than the value 2×10^–7 Nm^–1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere.

The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.

(b) When the current is flowing from south to north,

\(\theta\) = 0^o

f = 0

Hence there is no force on the conductor.

Roget's Spiral for Attraction Between Parallel Currents

Roget's Spiral for Attraction Between Parallel Currents

Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor \(\mu\). Hence it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1cm, the force per metre would be 5 × 10^–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable.

With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A.

Take a soft spring whose natural period of oscillations is about 0.5 – 1s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring, and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury.

Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply, and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit, and the cycle continues with tick, tick, tick,... In the beginning, you may require some small adjustments to get a good effect.

Keep your face away from mercury vapour as it is poisonous. Do not inhale mercury vapour for long.