Question

Consider the situation described in the previous problem. Suppose the current i enters the loop at the points A and leaves it at the point B. Find the magnetic field at the centre of the loop.

Answer 1

The loop ABCD can be considered as a circuit with two resistances in parallel, one along branch AB and other along branch ADC.

As, the sides of the loop are identical, their resistances are also same.

Let the resistance of each side be r.

The resistance of branch AB = r

The resistance of branch ADC = 3r

The current in the branches are calculated as:

iAB=i×3r3r+r=3i4

iADC=i×r3r+r=i4

As current follow the least resistive path so

Current in branch AB = 3i4

Current in branch ADC = i4

At the centre of the loop:

Magnetic field due to wire AD, DC and CB will be into the plane of paper according to right hand thumb rule.

Magnetic field due to wire AB will be out of the plane of paper according to right hand thumb rule.

Net magnetic field at the centre = BAD +BDC +BCB − BAB  which will be out of the plane of paper.

As, perpendicular distance of the centre from every wire will be equal to a2√ and angle made by corner points of each side at the centre is 45°.

BAD = μ0(i4)4π(a2√)(sin45°+sin45°)= μ0i8πaBAD = BDC=BCBBAD + BDC+BCB

Answer 2

Explanation:

To find the magnetic field at the centre of the loop:

Step 1:

The ABCD loop may be viewed as a circuit with two parallel resistors, one along the AB branch and the other along the ADC branch.

As the loop sides are identical, so are their resistances.  

Let the resistance be r on both sides.The resistance of branch AB = r

Branch resistance ADC = 3r

Step 2:

In the branches the current is calculated as:

$i_A_B = i\times \frac{3r}{3r+r} =\frac{3i}{4}$

$i_A_D_C = i \times \frac{3r}{(3r+r)} = \frac{i}{4}$

Follow the least resistive path as it is now.

Branch Current AB = $\frac{3i}{4}$  

Branch Current ADC = $\frac{i}{4}$

Step 3:

Centered around the loop:

According to the law of the right hand thumb, the magnetic field due to wire AD, DC and CB will be inside the paper plane.

By right-hand thumb law, the magnetic field due to wire AB will be out of the paper plane.  

Net magnetic field at center =$B_A_D + B_D_C + B_C_B -B_A_B$ that is out of the paper plane.  

Thus, the perpendicular distance of the core from each wire is equal to that of   a/√2 and a corner point angle at the center of each side is 45 °.

$BAD = \frac{( \mu_0 \frac{i}{4} )}{(4\pi\frac{a}{\sqrt{2}}} (sin45+sin45)$

$= \frac{\mu_0i}{8\pi a }$

$B_A_D = B_D_C= B_C_B$

$B_A_D + B_D_C+ B_C_B =B' = \frac{ 3\mu_0i}{8\pi a}$

$B_A_B = \frac{ ( \mu_0 ( \frac{3i}{4}) }{ (4\pi\frac{a}{\sqrt{2}} ) } (sin45 + sin45 )$

= $\frac{3 \mu_0i}{8\pi a}$

Step 4:

$B_n_e_t =B_' - B_A_B$

=0