Question

plzzzzz do it..............​

Answer 1

Answer:

The net magnetic field (B) is 2π mT

Given:-

Radius of both coil = 2 cm = 2 × 10⁻² m.

Each has 20 turns.

A carries 3A and B carries 4A.

Explanation:-

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Given, they are mutually perpendicular to each other, so the magnetic field due to both coils will be at right angles to each other irrespective of the current direction. Now, lets find the magnetic field by Coil A and B respectively. From the formula we know,

$\longrightarrow\sf B_{1}=\dfrac{\mu_{o}\;n\; I_{1}}{2\;R}$

Here,

n Denotes no. of turns.

I₁ Denotes current in coil A.

R denotes radius.

Substituting the values,

$\longrightarrow\sf B_{1}=\dfrac{4\;\pi\times 10^{-7}\times 20\times 3}{2\times 10^{-2}}\\\\\\\\\longrightarrow\sf B_{1}=\dfrac{4\;\pi\times 10^{-7}\times 10\times 3}{10^{-2}}\\\\\\\\\longrightarrow\sf B_{1}=\dfrac{120\;\pi\times 10^{-7}\times 10^{2}}{1}\\\\\\\\\longrightarrow\sf B_{1}=120\;\pi\times 10^{-5}\\\\\\\\\longrightarrow\sf B_{1}=1.20\;\pi\times 10^{-3}\\\\\\\\\longrightarrow\boxed{\sf B_{1}=1.2\;\pi\times 10^{-3}}\quad\dots\dots\sf (1)$

Now, let's find the magnetic field in coil B,

$\longrightarrow\sf B_{2}=\dfrac{\mu_{o}\;n\; I_{2}}{2\;R}$

Here,

n Denotes no. of turns.

I₂ Denotes current in coil B.

R denotes radius.

Substituting the values,

$\longrightarrow\sf B_{2}=\dfrac{4\;\pi\times 10^{-7}\times 20\times 4}{2\times 10^{-2}}\\\\\\\\\longrightarrow\sf B_{2}=\dfrac{4\;\pi\times 10^{-7}\times 10\times 4}{10^{-2}}\\\\\\\\\longrightarrow\sf B_{2}=\dfrac{160\;\pi\times 10^{-7}\times 10^{2}}{1}\\\\\\\\\longrightarrow\sf B_{2}=160\;\pi\times 10^{-5}\\\\\\\\\longrightarrow\sf B_{2}=1.60\;\pi\times 10^{-3}\\\\\\\\\longrightarrow\boxed{\sf B_{2}=1.6\;\pi\times 10^{-3}}\quad\dots\dots\sf (2)$

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As we discussed earlier Both B₁ and B₂ are perpendicular to each other so, the net magnetic field will be given by,

$\longrightarrow\sf B=\sqrt{\bigg(B_{1}\bigg)^{2}+\bigg(B_{2}\bigg)^{2}}$

B denotes Net magnetic field, Substituting the values,

$\longrightarrow\sf B=\sqrt{\bigg(1.2\;\pi\times 10^{-3}\bigg)^{2}+\bigg(1.6\;\pi\times 10^{-3}\bigg)^{2}}\\\\\\\\\longrightarrow\sf B=\pi\times10^{-3}\sqrt{\bigg(1.2\bigg)^{2}+\bigg(1.6\bigg)^{2}}\\\\\\\\\longrightarrow\sf B=\pi\times 10^{-3}\sqrt{\bigg(1.44\bigg)+\bigg(2.56\bigg)} \\\\\\\\\longrightarrow\sf B=\pi\times 10^{-3}\sqrt{\bigg(4.00\bigg)}\\\\\\\\\longrightarrow\sf B=\pi\times 10^{-3}\times 2\\\\\\\\\longrightarrow\sf B=2\;\pi\times 10^{-3}\\\\\\\\\longrightarrow\sf B=2\;\pi\;mT$

$\longrightarrow\large{\underline{\boxed{\red{\sf B=2\;\pi\; mT}}}}$

∴ The net magnetic field (B) is 2π mT.

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Answer 2

Answer:

Where the two coils are kept perpendicular to each other then resultant field in one coil is directed along the axis of coil and the other downwards, hence resultant field is

${b}^{1 } \: = \sqrt{ {b}^{2} + {b}^{2} } = \sqrt{2b}$

Ratio of magnetic field due to one coil and resultant is

$\frac{b}{\sqrt[b]{2 } } = \frac{1}{ \sqrt{2} }$

hope this helps you