A wire <math><mi>A</mi></math> , bent in the shape of an arc of a circle, carrying a current of <math><mn>2</mn><mtext></mtext><mi mathvariant="normal">A</mi></math> and having radius <math><mn>2</mn><mtext></mtext><mi mathvariant="normal">c</mi><mi mathvariant="normal">m</mi></math> and another wire <math><mi>B</mi></math> , also bent in the shape of arc of a circle, carrying a current of <math><mn>3</mn><mtext></mtext><mi mathvariant="normal">A</mi></math> and having radius of <math><mn>4</mn><mtext></mtext><mi mathvariant="normal">c</mi><mi mathvariant="normal">m</mi></math> , are placed as shown in the figure. The ratio of the magnetic fields due to the wires <math><mi>A</mi></math> and <math><mi>B</mi></math> at the common centre <math><mi>O</mi></math> is:

BA=μ0Iθ4πR⇒BABB=IAθARBIBθBRA⇒23π2 (4)35π3 [2]⇒65 .

A wire&nbsp;<math><mi>A</mi></math>&nbsp;, bent in the shape of an arc of a circle, carrying a current of&nbsp;<math><mn>2</mn><mtext></mtext><mi mathvariant="normal">A</mi></math>&nbsp;and having radius&nbsp;<math><mn>2</mn><mtext></mtext><mi mathvariant="normal">c</mi><mi mathvariant="normal">m</mi></math>&nbsp;and another wire&nbsp;<math><mi>B</mi></math>&nbsp;, also bent in the shape of arc of a circle, carrying a current of&nbsp;<math><mn>3</mn><mtext></mtext><mi mathvariant="normal">A</mi></math>&nbsp;and having radius of&nbsp;<math><mn>4</mn><mtext></mtext><mi mathvariant="normal">c</mi><mi mathvariant="normal">m</mi></math>&nbsp;, are placed as shown in the figure. The ratio of the magnetic fields due to the wires&nbsp;<math><mi>A</mi></math>&nbsp;and&nbsp;<math><mi>B</mi></math>&nbsp;at the common centre&nbsp;<math><mi>O</mi></math>&nbsp;is:

Ask & Answer Question

Option 1 -

$6 : 5$

Option 2 -

$2 : 5$

Option 3 -

$6 : 4$

Option 4 -

$4 : 6$

5 Views | Posted 4 months ago

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

Correct Option - 1

Detailed Solution:

$B_{A} = \frac{μ_{0} I θ}{4 π R}$

$\Rightarrow \frac{B_{A}}{B_{B}} = \frac{I_{A} θ_{A} R_{B}}{I_{B} θ_{B} R_{A}}$

$\Rightarrow \frac{2 (\frac{3 π}{2}) (4)}{3 [\frac{5 π}{3}] [2]}$

$\Rightarrow \frac{6}{5}$ .

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