Magnetic Forces Between Two Parallel Current Carrying Conductors Questions

1.0Nature of Force

• Two long parallel conductors carrying currents in the same direction attract each other.
• Two long parallel conductors carrying currents in the opposite direction repel each other.

2.0Magnitude of Force

• The net magnetic force acts on a current carrying conductor due to its own field is zero. So consider two infinite long parallel conductors separated by distance 'd' carrying currents I₁ and I₂.
• Magnetic field at each point on conductor (2) due to current I₁ is

$B_1=\frac{{\mu }_0{I}_1}{2\pi D}$ (uniform magnetic field for conductor 2)

• Magnetic field at each point on conductor (i) due to current I₂ is

$B_2=\frac{{\mu }_0{I}_2}{2\pi D}$ (uniform magnetic field for conductor 1)

• Considering a small element of length \text { 'dl' } on each conductor. These elements are right angle to the external magnetic field, so magnetic force experienced by elements of each conductor given as 

$d{F}_{12}=B_2{1}_{1}dl=\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)I_{1}dl$………(1)    (where $I_{1}dl\perp B_2$)

$d{F}_{21}=B_1{1}_{2}dl=\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)I_{2}dl$………(1)    (where $I_{2}dl\perp B_1$)

Where d F_{12} is a magnetic force on element of conductor(1) due to field of conductor(2) and d F_{21} is a magnetic force on element of conductor(2),due to field of conductor (1)

3.0Magnitude Force Per Unit Length of Each Conductor

$\frac{d{F}_{12}}{dl}=\frac{d{F}_{21}}{dl}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$

• In SI, $f=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}N/m$
• In CGS, $f=\frac{2I_1{I}_2}{d}\text{ dyne }/\mathrm{cm}$
• Force scale $f=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$ is applicable when at least one conductor must be of infinite length, so it behaves like a source of uniform magnetic field for other conductors
• Magnetic Force on conductor ‘LN’ is = $f\times l\Rightarrow F_{LN}=\left(\frac{{\mu }_0{I}_1{I}_2}{2\pi d}\right)l$

4.0Definition of Ampere

• Magnetic force/unit length for both infinite length conductor gives as 

$f=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\frac{\left(4\pi \times 10^{-7}\right)(1)(1)}{2\pi (1)}=2\times 10^{-7}\mathrm{N}/\mathrm{m}$

• 'Ampere is the current which, when passed through each of two parallel infinite long straight conductors placed in free space at a distance of 1 m from each other, produces between them a force of 2 × 10^–7 N/m

5.0Equilibrium of Free Wire

Case I : Upper wire is free: Consider a long horizontal wire which is rigidly fixed; another wire is placed directly above and parallel to fixed wire.    

• Magnetic Force per unit length of free wire $f_m=\frac{{\mu }_0{I}_1{I}_2}{2\pi h}$, and it is repulsive in nature because currents are unlike.
• Free wire may remain suspended if the magnetic force per unit length is equal to weight of its unit length.
• At balanced condition $f_m=W^{\prime }$
• Weight per unit length of free wire=$\frac{{\mu }_0{I}_1{I}_2}{2\pi h}=\frac{m}{l}g$ (stable equilibrium condition)

Case II : Lower wire is free : Consider a long horizontal wire which is rigidly fixed. Another wire is placed directly below and parallel to the fixed wire.

• Magnetic force per unit length of free wire is $f_m=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$, and it is attractive in nature because current is like.
• Free wire may remain suspended if the magnetic force per unit length is equal to weight of its unit length.
• At balanced condition $f_m=W^{\prime }$
• Weight per unit length of free wire= $\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\frac{m}{l}g$   (unstable equilibrium condition)

6.0Sample Questions on Magnetic Forces Between Two Parallel Current Carrying Conductors Questions

Q-1.Two long and parallel wires are at a separation of 0.1 m and a current of 5 A is gliding in each of these wires. The force per unit length due to these parallel wires will be.

Solution: $F=\frac{{\mu }_0}{4\pi }\frac{2I_1{I}_2}{a}=\frac{10^{-7}\times 2\times 5\times 5}{0.1}=5\times 10^{-5}\mathrm{N}/\mathrm{m}$

Q-2. A current of 5A flows through two long parallel wires. The magnetic force on each wire is 4 × 10^–3 N/m. If their currents make half and separation between them makes double then magnetic force per unit length of each wire becomes.

Solution: Magnetic force per unit length 

$f_m=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$…………..(1)

$f_m^{\prime }=\frac{{\mu }_0\left(\frac{I_1}{2}\right)\left(\frac{I_2}{2}\right)}{2\pi (2d)}=\frac{{\mu }_0{I}_1{I}_2}{2d}\times \frac{1}{8}$………(2)

From (1) and (2)

$f_m^{\prime }=\frac{f_m}{8}=\frac{4\times 10^{-3}}{8}=0.5\times 10^{-3}\mathrm{N}/\mathrm{m}$

Q-3.Three long, straight and parallel wires carrying currents are arranged as shown in figure. The force experienced by 10 cm length of wire Q is

Solution: Force on wire Q due to wire P is

$F_P=10^{-7}\times \frac{2\times 30\times 10}{0.1}\times 0.1=6\times 10^{-5}N(\text{ Toward Left })$

Force on wire Q due to wire R is 

$F_R=10^{-7}\times \frac{2\times 20\times 10}{0.02}\times 0.1=20\times 10^{-5}N(\text{ Toward Right })$

Hence $F_{net}=F_R-F_P=14\times 10^{-5}\mathrm{N}(\text{ Towards Right })$

Q-4.What is the net force on the square coil

Solution: Force on side BC and AD are equal but opposite so their net will be zero.

But

$F_{AB}=10^{-7}\times \frac{2\times 2\times 1}{2\times 10^{-2}}\times 15\times 10^{-2}=3\times 10^{-6}\mathrm{N}$

And $F_{CD}=10^{-7}\times \frac{2\times 2\times 1}{\left(12\times 10^{-2}\right)}\times 15\times 10^{-2}=0.5\times 10^{-6}\mathrm{N}$

$F_{net}=F_{AB}-F_{CD}=2.5\times 10^{-6}\mathrm{N}=25\times 10^{-7}\mathrm{N},(\text{ Towards the wire })$

Q-5.A long horizontal wire is rigidly fixed and carries 100A current. Another wire of linear mass density 2 × 10^–3 kg/m placed below and parallel to the fixed wire. If the free wire kept 2 cm below and hangs in air, then current in free wire is:-

Solution: At balanced condition of free wire :-

$f_m=\lambda g$

$\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\lambda g$

$I_2=\frac{2\pi \lambda gd}{{\mu }_0{I}_1}=\frac{2\times 3.14\times 2\times 10^{-3}\times 9.8\times 2\times 10^{-2}}{4\pi \times 10^{-7}\times 100}=19.6\mathrm{A}\text{ (in Free Wire) }$