Question

an equilateral triangular loop is made up of a wire of uniform resistance a current I enters through one of the vertices of the triangle and exit from other vertices of triangle of side a the magnitude of magnetic field at the centre of equilateral loop is​

Answer 1

Given :

Side length of the equilateral triangle formed by the wire = $l$

Magnitude of current entering through one of the vertices of the triangle = $I$

To Find :

Magnitude of magnetic field at the centre of the equilateral loop = ?

Solution :

Since the wire is of uniform resistance ,so let resistance of each side of triangular loop be = r

Now the current $I$ entering through 1st vertex divides into two sides as$I_1$ and $I_2$ and hence the current through the 3rd side is

And  the connection looks like a parallel connection , so the currently will be inversly proportional to the resistance for each branch .

So we can write :

$\frac{I_1}{I_2} = \frac{r_2 }{r_1}$

Or , $\frac{I_1}{I_2} = \frac{2r }{r_}$

Or, $I_1 = 2 I_2$    - (1)

∴$I_1 +I_2 = I$  -(2)

Putting the value of $I_2$ in eq 2 we get :

$I_1 +\frac{I_1}{I_2}= I$

Or , $3I_1 = 2I$

So , we get:

$I_1 = \frac{2I}{3}$ and $I_2 = \frac{I_1}{2} = \frac{I}{3}$

Now we know that magnetic field due to a straight current carrying wire at a distance 'd' from wire is given as :

$B = \frac{\mu_0I}{4\pi d} (sin\theta _1 - sin\theta _2)$

And for an equilateral triangle the values of $\theta _1$ and $\theta_2$ is 60° and (-60°) respectively .

And d which is distance of centre from one side is = $\frac{a}{2\sqrt{3} }$

So , magnetic field due to one side of the triangle at the centre is :

$B_1 = \frac{\mu_0 I_1}{2 \pi \frac{a}{2\sqrt{3} } } (sin60 - sin(-60))$

    $= \frac{\mu_0 I_1 \sqrt3}{\frac{\pi a}{\sqrt3} }$

    $= \frac{\mu_0I_1 \times 3}{\pi a}$

And magnetic field due to other two sides which are in series is :

$B_2 = \frac{\mu_0 I_2 }{2\pi \frac{a}{2\sqrt{3} } }(sin60 - sin(-60))$

    =$=\frac{\mu _0 I_2 \times 3}{\pi a}$

Taking vector sum ,

$B_1$ is downward and $B_2$ is upward for both sides , so :

$B_{net}=2\times B_2 - B_1$

       $=\frac{2 \ mu_0 I_2 \times 3 }{\pi a } - \frac{\mu_0 I_1\times 3}{\pi a }$

       $= \frac{3 \mu_0}{\pi a } (2I_2- I_1)$

      ∴$2I_2 = I_1$

So, $B_{net} = 0$

So, net magnetic field at the centre of the triangular loop is 0 .