Question

figure shows square loop made from a uniform wire .the current is made to flow through the loop.the net magnetic field at the centre O of loop is
​

Answer 1

Answer:

As the material of the loop is same, resistance through the path with 3 sides will be 3 times the resistance through the path with one side. So current will get divided in the ratio shown in the figure.

Also, θ

1

and θ

2

mentioned in the formula will be equal to 45

∘

as the loop is square loop. ( can be deduced from the figure)

The magnetic field due to the shorter path will be

4π.

2

a

μ

0

.0.75I

The magnetic field due to the longer path will be 3×

4π.

2

a

μ

0

.0.25I

=

4π.

2

a

μ

0

.0.75I

By Ampere's right hand grip rule, we can conclude that the magnetic field due to both these paths is opposite in direction. So, the fields will cancel out and the resultant field at the center will be zero.

Answer 2

To find:

Net magnetic field intensity at the centre of the loop.

Diagram:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(2,2){\vector(0, 1){2}}\put(2,4){\vector(1,0){2}}\put(4,4){\vector(0,-1){2}}\put(4,2){\vector(-1,0){2}}\put(4.5,3.5){\vector(0,1){0.5}}\put(4.5,2.5){\vector(0,-1){0.5}}\put(4.5,3){s}\put(2.85,3){ \times }\put(3,3.5){ B_{net} }\put(3,4.25){i}\end{picture}}$

Calculation:

The magnetic field intensity at the centre of the square loop will be vector addition of the magnetic field intensity due to individual segments of the loop.

$B_{net} = B1 + B2 + B3 + B4$

$= > B_{net}= B + B + B + B$

$= > B_{net}= 4B$

$= > B_{net}= 4 \bigg \{ \dfrac{ \mu_{0}i}{4\pi d} ( \sin {45}^{ \circ } + \sin {45}^{ \circ} ) \bigg \}$

$= > B_{net}= 4 \bigg \{ \dfrac{ \mu_{0}i}{4\pi d} ( 2\sin {45}^{ \circ }) \bigg \}$

$= > B_{net}= 4 \bigg \{ \dfrac{ \mu_{0}i}{4\pi d} ( 2 \times \dfrac{1}{ \sqrt{2} } ) \bigg \}$

$= > B_{net}= \dfrac{ \sqrt{2} \mu_{0}i}{\pi d}$

$= > B_{net}= \dfrac{ \sqrt{2} \mu_{0}i}{\pi (\frac{s}{2} )}$

$= > B_{net}= \dfrac{ 2 \sqrt{2} \mu_{0}i}{\pi s}$

So, final answer is:

$\boxed{ \sf{B_{net}= \dfrac{ 2 \sqrt{2} \mu_{0}i}{\pi s} }}$