Question

A wire is converted in form of square of side
20 cm. If current in it is 10 A then magnetic
field at centre will be :-
(1) 22 x 104T
(2) 42 10T
(3) 8 V2 x 10 T
(4) zero​

Answer 1

Answer:

that the AI system remains unbiased.

1) project review

2) system tuning

3) evaluation and deployment

4) model selection

Answer 2

The given options are wrong but refer the solution below, it is definitely correct.

The law that will be used here is Biot-Savart's law.

Given,

side of square=20 cm

side in m=20/100

side of square=0.2 m

current flowing through the square=10 A.

To find,

magnetic field at the centre.

Solution:

according to biot-savart's law,

$B=\frac{nu}{4\pi } \frac{I}{a}$

where,

B-magnetic field

nu-relative permeability    $(\frac{nu}{4\pi } =10^{-7} )$

I-current

a-length

B due to one side,

$B=\frac{nu}{4\pi } \frac{I}{a} (sin 45 degree+ sin 45 degree)\\B=\frac{nu}{4\pi } \frac{I}{a}(\frac{1}{\sqrt{2} } +\frac{1}{\sqrt{2} })\\B=\frac{nu}{4\pi } \frac{I}{a}(\frac{2}{\sqrt{2} } )\\B=\frac{nu}{4\pi } \frac{I}{a}\sqrt{2} (inwards)$

Bnet due to all four sides (magnetic field at centre),

$Bnet=4XB\\Bnet=4X\frac{nu}{4\pi } \frac{I}{a} \sqrt{2} \\Bnet=4X\sqrt{2} X\frac{nu}{4\pi } \frac{I}{a} \\Bnet=4X\sqrt{2}X10^{-7} X\frac{10}{0.2} \\Bnet=282.8X10^{-7} T$

The magnetic field at the centre of the square is $282.8X10^{-7} T$.

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