Question

ABCD is a rectangular loop made of uniform wire. If $AD = BC = 2$ cm, what is the magnetic force per unit length acting on wire $DC$ due to wire $AB$ if ammeter reads $20 A$. (The length of $AB$ and $DC$ are very-very large in comparison with other two sides $AD$ and $BC$) (Assume $P$ and $Q$ are mid-point of $AD$ and $BC$ respectively)

Answer 1

1 × 10^{-3} N/m

Answer 2

The problem asks for the magnetic force per unit length acting on wire DC due to wire AB in a rectangular loop ABCD.

1. Determine the current in wires AB and DC:

The total current flowing into the loop from the ammeter is $I_{total} = 20$ A.

The current enters at point P (midpoint of AD) and exits at point Q (midpoint of BC).

Since the loop is made of uniform wire and P and Q are midpoints, the current splits equally into two parallel paths: PABQ and PDCQ.

The resistance of path PABQ is $R_{PABQ} = R_{PA} + R_{AB} + R_{BQ}$.

The resistance of path PDCQ is $R_{PDCQ} = R_{PD} + R_{DC} + R_{CQ}$.

Given that P and Q are midpoints, $PA = PD = AD/2$ and $BQ = CQ = BC/2$.

Since ABCD is a rectangle, $AD = BC = 2$ cm. So, $PA = PD = BQ = CQ = 1$ cm.

The lengths of AB and DC are stated to be very large and equal ($AB = DC$).

Since the wire is uniform, the resistance per unit length ($\rho$) is constant.

Thus, $R_{PA} = R_{PD} = \rho \times 1 \text{ cm}$, $R_{BQ} = R_{CQ} = \rho \times 1 \text{ cm}$, and $R_{AB} = R_{DC} = \rho \times L_{AB}$.

Therefore, $R_{PABQ} = \rho(1 + L_{AB} + 1) = \rho(L_{AB} + 2)$ and $R_{PDCQ} = \rho(1 + L_{DC} + 1) = \rho(L_{DC} + 2)$.

Since $L_{AB} = L_{DC}$, we have $R_{PABQ} = R_{PDCQ}$.

As the resistances of the two paths are equal, the current splits equally: $I_{AB} = I_{DC} = \frac{I_{total}}{2} = \frac{20 \text{ A}}{2} = 10 \text{ A}$.

The current in wire AB flows from A to B, and the current in wire DC flows from D to C. These currents are in the same direction (e.g., both rightward).

2. Identify the distance between the wires:

The distance between wire AB and wire DC is the side length AD (or BC).

Given $d = AD = 2 \text{ cm} = 0.02 \text{ m}$.

3. Apply the formula for magnetic force per unit length:

For two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$, the magnetic force per unit length ($F/L$) is given by:

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

where $\mu_0 = 4 \pi \times 10^{-7} \text{ T} \cdot \text{m/A}$ (permeability of free space).

4. Calculate the force per unit length:

Substitute the values: $I_1 = I_{AB} = 10 \text{ A}$, $I_2 = I_{DC} = 10 \text{ A}$, and $d = 0.02 \text{ m}$.

$F/L = \frac{(4 \pi \times 10^{-7} \text{ N/A}^2) \times (10 \text{ A}) \times (10 \text{ A})}{2 \pi \times (0.02 \text{ m})}$

$F/L = \frac{2 \times 10^{-7} \times 100}{0.02} \text{ N/m}$

$F/L = \frac{2 \times 10^{-5}}{2 \times 10^{-2}} \text{ N/m}$

$F/L = 1 \times 10^{-3} \text{ N/m}$

Since the currents are in the same direction, the force is attractive (wire DC is attracted towards wire AB).

The final answer is $1 \times 10^{-3} N/m$.

Explanation of the solution:

1. The 20 A current splits equally into two parallel paths (AB and DC) because the wire is uniform and the entry/exit points (P, Q) are midpoints, making the resistance of both paths equal. So, $I_{AB} = I_{DC} = 10$ A.

2. The distance between the parallel wires AB and DC is given as $AD = 2$ cm = 0.02 m.

3. The magnetic force per unit length between two long parallel current-carrying wires is calculated using the formula $F/L = \frac{\mu_0 I_1 I_2}{2 \pi d}$.

4. Substituting the values ($I_1 = 10$ A, $I_2 = 10$ A, $d = 0.02$ m, $\mu_0 = 4\pi \times 10^{-7}$ N/A$^2$), the force per unit length is $1 \times 10^{-3}$ N/m.