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Magnetic Field due to a Current — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMagnetic Field due to a Current5 Marks
Question
A square loop PQRS carrying a current of 6.0A is placed near a long wire carrying 10A as shown in figure.
  1. Show that the magnetic force acting on the part PQ is equal and opposite to that on the part RS.
  2. Find the magnetic force on the square loop.
✓

Answer


$\text{I}_2=6\text{A},\ \text{I}_1=10\text{A}$
  1. $\text{F}_\text{PQ}$
‘F’ on $=\frac{\mu_0\times30}{\text{x}}\int_1\frac{\text{dx}}{\text{x}}=30\times4\times10^{-7}\times[\log\text{x}]^2_1$
$=120\times10^{-7}[\log3-\log1]$
Similarly force of $\overrightarrow{\text{F}}_\text{RS}=120\times10^{-7}[\log3-\log1]$
  1. So, $\overrightarrow{\text{F}}_\text{PQ}=\overrightarrow{\text{F}}_\text{RS}$
$\overrightarrow{\text{F}}_\text{PS}=\frac{\mu_0\times\text{i}_1\text{i}_2}{2\pi\times1\times10^{-2}}-\frac{\mu_0\times\text{i}_1\text{i}_2}{2\pi\times2\times10^{-32}}$
$=\frac{2\times6\times10\times10^{-7}}{10^{-2}}-\frac{2\times10^{-7}\times6\times6}{2\times10^{-2}}=8.4\times10^{-4}\text{N}$ (Towards right)
$\overrightarrow{\text{F}}_\text{RQ}=\frac{\mu_0\times\text{i}_1\text{i}_2}{2\pi\times3\times10^{-2}}-\frac{\mu_0\times\text{i}_1\text{i}_2}{2\pi\times2\times10^{-32}}$
$=\frac{4\pi\times10^{-7}\times6\times10^{-2}}{2\pi3\times10^{-2}}-\frac{4\pi\times10^{-7}\times6\times6}{2\pi\times2\times10^{-2}}$
$=4\times10^{-4}+36\times10^{-5}=7.6\times10^{-4}\text{N}$
Net force towards down
$= (8.4 + 7.6) × 10^{-4} = 16 × 10^{-4} \text{N}$

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