Question

Four wires carrying current I1=2A, I2=4A, I3=6A and I4=8A respectively cut the page perpendicularly as shown in the figure. The value of B . dl for the loop shown would be

Answer 1

4μ₀

Answer 2

The problem asks us to find the value of the line integral of the magnetic field $\vec{B}$ around a closed loop, $\oint \vec{B} \cdot d\vec{l}$. This can be directly calculated using Ampere's Circuital Law.

Ampere's Circuital Law:

Ampere's Circuital Law states that the line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_0$ times the total current enclosed by that path. Mathematically, this is expressed as:

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$$

where:

• $\oint \vec{B} \cdot d\vec{l}$ is the line integral of the magnetic field around the closed loop.
• $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \text{ T m/A}$).
• $I_{enclosed}$ is the net current passing through the area enclosed by the loop.

Step-by-step calculation:

1. Identify the currents and their directions:

   • $I_1 = 2A$ (out of the page, represented by a dot)
   • $I_2 = 4A$ (into the page, represented by a cross)
   • $I_3 = 6A$ (out of the page, represented by a dot)
   • $I_4 = 8A$ (into the page, represented by a cross)

2. Determine which currents are enclosed by the loop:

   From the figure, the closed loop clearly encloses currents $I_1$, $I_2$, and $I_3$. Current $I_4$ is outside the loop and therefore does not contribute to $I_{enclosed}$.

3. Establish the sign convention for enclosed currents:

   The direction of integration along the loop is indicated by an arrow, which is counter-clockwise. According to the right-hand thumb rule, if you curl the fingers of your right hand in the direction of the loop (counter-clockwise), your thumb points out of the page. Therefore, currents coming out of the page are considered positive, and currents going into the page are considered negative.

   Applying this convention:

   • $I_1 = +2A$ (out of the page)
   • $I_2 = -4A$ (into the page)
   • $I_3 = +6A$ (out of the page)

4. Calculate the net enclosed current ($I_{enclosed}$):

   $$I_{enclosed} = I_1 + I_2 + I_3$$ $$I_{enclosed} = 2A + (-4A) + 6A$$ $$I_{enclosed} = 2A - 4A + 6A$$ $$I_{enclosed} = 4A$$

5. Apply Ampere's Circuital Law:

   Substitute the value of $I_{enclosed}$ into Ampere's Law:

   $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$$ $$\oint \vec{B} \cdot d\vec{l} = \mu_0 (4A)$$ $$\oint \vec{B} \cdot d\vec{l} = 4\mu_0$$

The value of $\oint \vec{B} \cdot d\vec{l}$ for the given loop is $4\mu_0$.

The final answer is $\boxed{4\mu_0}$.

Explanation of the solution:

Ampere's Circuital Law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$, is applied. Identify currents enclosed by the loop ($I_1, I_2, I_3$). Use the right-hand rule with the given counter-clockwise loop direction to assign signs: currents out of page are positive, into page are negative. Calculate net enclosed current: $I_{enclosed} = 2A - 4A + 6A = 4A$. Substitute into Ampere's Law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 (4A) = 4\mu_0$.