Q.Two long and parallel conductors A and B are carrying currents 8.0 A and 5.0 A in the same direction. A and B are separated by a distance of 4.0 cm. Calculate the force on a length of 1 cm of wire A.

Two parallel current-carrying wires attract if their currents are in the same direction; the force per unit length is mu_0I1I2/(2pid).

Each current-carrying wire produces a magnetic field around it, and this field exerts a force on the other current-carrying wire (F = I*L x B). The standard result for the force per unit length between two long straight parallel wires carrying currents I1 and I2, separated by a distance d, is

F/L = mu_0 * I1 * I2 / (2 * pi * d)

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Both statements are individually correct, but the field being zero at the midpoint is not why the two wires attract each other.

Checking the Assertion: Two infinitely long straight parallel conductors carrying current in the SAME direction do attract each other. Each wire sits in the magnetic field created by the other wire, and using $\vec F = I\vec L \times \vec B$ (or the right-hand/Fleming's left-hand rule), the force on each wire due to the other's field points towards the other wire. So the Assertion is TRUE.

Checking the Reason: Take the two wires along the $y$-axis at $x=-a$ and $x=+a$, both carrying current $I$ in the $+y$ direction. At the midpoint (origin), using $\vec B = \dfrac{\mu_0 I}{2\pi r}\hat\phi$ with $\hat\phi = \hat I \times \hat r$: the field due to the left wire points in $+\hat y'$s perpendicular direction (say $+\hat z$), while the field due to the right wire (displacement now in $-\hat x$ from that wire) points in the opposite transverse direction ($-\hat z$). Since both wires are equidistant and carry equal current, these two fields are equal in magnitude and opposite in direction — they cancel exactly. So the net field at the midpoint IS zero when the currents …

Using the force per unit length between two wires, μ₀ works out to dimensions [M L T⁻² A⁻²].

The force per unit length between two parallel current-carrying wires is

$\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2\pi d}$

Solving for μ₀:

$\mu_0 = \frac{2\pi d\,(F/\ell)}{I_1 I_2}$

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Adjacent turns of a current-carrying spring act like parallel wires carrying current in the same direction, which attract each other by the magnetic force between parallel currents — so the coils are pulled together.

Two straight parallel conductors carrying currents in the SAME direction attract each other (force per unit length $F/l = \mu_0 I_1 I_2 / 2\pi d$, attractive for like-directed currents, repulsive for opposite). A spring is essentially a coil of many closely-spaced turns; each turn carries current in the same sense as its neighbours. Treating adjacent turns as parallel current-carrying wires, they attract each other, so the spring's turns are pulled closer together ( …

Connected in series, the two freely suspended parallel wires carry equal currents in opposite directions — the current goes out along one wire and returns along the other. Antiparallel currents repel, so the wires move apart. Both statements are true and the Reason is exactly why the wires separate. The correct option is (A).

Like charges repel electrostatically; this force exceeds the magnetic attraction at ordinary speeds, so the beams diverge.

Two parallel electron beams experience two effects:

• As parallel currents in the same direction, the magnetic force is attractive.
• As streams of like (negative) charges, the electrostatic force is repulsive. …

Two parallel current-carrying wires attract each other (same direction) with a force per unit length given by mu0 I1 I2 / (2pid).

The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$, separated by distance $d$, is

$\dfrac{F}{l} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$

Substituting $\mu_0 = 4\pi\times10^{-7}\ \text{T m/A}$, $I_1 = I_2 = 1\ \text{A}$, $d = 1\ \text{m}$:

$\dfrac{F}{l} = \dfrac{4\pi\times10^{-7}\times 1\times 1}{2\pi\times 1} = 2\times10^{-7}\ \text{N/m}$

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Two parallel current-carrying conductors carrying current in the same direction attract each other.

Each current-carrying conductor sets up a magnetic field around itself (by the Biot-Savart/Ampere law), and the other conductor, carrying current in that field, experiences a force $F = BIL$ (via $\vec F = I\vec L\times\vec B$). Working out the directions with the right-hand rule shows that when the currents flow in the same direction, the force on each conductor points toward the other - i.e. the conductors …

Two parallel current-carrying wires exert a magnetic force on each other; the force per unit length is F/l = (mu0 I1 I2)/(2 pi d), and it is attractive when the currents flow in the same direction.

The magnetic field produced by wire 1 at the location of wire 2 (distance d away) is:

B1 = (mu0 I1)/(2 pi d)

This field exerts a force per unit length on wire 2 (carrying current I2):

F/l = B1 * I2 = (mu0 I1 I2)/(2 pi d)

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Like (same-direction) currents attract; unlike (opposite-direction) currents repel.

Each current-carrying conductor sets up a magnetic field, and the second conductor carrying current experiences a force in that field (F = BIL).

• Same direction: Using the right-hand rule, the field of one wire at the location of the other, combined with F = I L × B, gives a force pulling the wires toward each other. So the force is attractive.
• Opposite directions: the same analysis reverses the force direction, pushing the wires apart. So the force is repulsive. …