Question

Equal current I flows in two segments of a circular loop in the direction shown in figure. radius of the loop is “a”.magnetic field at the centre of the loop is?

Answer 1

Answer:answer can be get from the formula of circular arc

Explanation:

Answer 2

Concept:

A field of force could be a vector field that describes the magnetic influence on moving charge, electric currents, and magnetic materials. A moving charge in a very force field experiences a force perpendicular to its own velocity and to the field of force. Magnetic fields are represented using magnetic lines. it's a visible tool wont to visualize the direction.

Given:

We are only if equal current $I$ flows in two segments of a circular loop within the direction shown within the figure. the radius of the loop is $a.$

Find:

We have to seek out the magnetic flux at the middle of the loop.

Solution:

As we all know that flux of circular loop is given by,

$B=\frac{\mu_{0}I}{2R}$

Here, $I$ is electrical current, $R$ is Radius of circular loop

Magnetic field due to $ADB$ is $B_{1}=\left(\frac{\theta}{2\pi}\right)\frac{\mu_{0}I}{2R} .......(1)$

Perpendicular to paper outwards.

And magnetic field due to $ACB$ is $B_{2}=\left(\frac{2\pi-\theta}{2\pi}\right)\frac{\mu_{0}I}{2R} .........(2)$

Perpendicular to paper inwards

Net magnetic field of force is given by $B_{net}=B_2-B_1$

Put the worth in net field of force from equation (1) and equation (2), we get

$B_{net}=\left(\frac{2\pi-\theta}{2\pi}\right)\frac{\mu_{0}I}{2R}-\left(\frac{\theta}{2R}\right)\frac{\mu_{0}I}{2R}$

Now, we'll simplify the above expression, we get

$\begin{aligned}B_{net}&=\left(\frac{2\pi-\theta}{2\pi}-\frac{\theta}{2\pi}\right)\frac{\mu_{0}I}{2R}\\ &=\left(\frac{2\pi-2\theta}{2\pi}\right)\frac{\mu_{0}I}{2R}\\ &=\left(\frac{\pi-\theta}{\pi}\right)\frac{\mu_{0}I}{2R}\end$

Hence, the magnetic field at the middle of the loop is$\left(\frac{\pi-\theta}{\pi}\right)\frac{\mu_{0}I}{2R}$.

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