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Magnetic Field due to a Current Element Physics Moving Charges and Magnetism Physics Class 12th

AB µ r² long straight wire with a circular cross-section having radius R, is carrying a steady current I: The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be:

Option 1 -

B µ r²

Option 2 -

B µ r

Option 3 -

B µ $\frac{1}{r^{2}}$

Option 4 -

B µ $\frac{1}{r}$

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a month ago

Correct Option - 2

Detailed Solution:

for $π R^{2} A r e a \to' l'$ current\

1 unit Area ® $\frac{l}{π R^{2}}$

$π r^{2} \to \frac{l}{π R^{2}} \times π r^{2}$

$i = \frac{l r^{2}}{R^{2}}$

Now, consider Amperian loop of radius small ‘r’ ln Amperian loop magnetic field will be tangential to the amperian loop.

$? \vec{B} . d \vec{i} = μ_{0} l_{e n c l o s e d}$ (Ampere circuital law)

$B = \frac{μ_{0}}{2 π} \frac{l}{R^{2}} r$

$B \propto r$

<p>for <span contenteditable="false"> <math> <mrow> <mi>π</mi> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mtext> </mtext> <mtext> </mtext> <mi>A</mi> <mi>r</mi> <mi>e</mi> <mi>a</mi> <mo>→</mo> <mo>'</mo> <mi>l</mi> <mo>'</mo> </mrow> </math> </span> current\</p><div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1757388392phpVuYrnw.jpeg" alt="" width="267" height="93" loading="lazy"></picture></div></div><p>1 unit Area ®<span contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mi>l</mi> </mrow> <mrow> <mi>π</mi> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> </span> </p><p>              <span contenteditable="false"> <math> <mrow> <mi>π</mi> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>→</mo> <mfrac> <mrow> <mi>l</mi> </mrow> <mrow> <mi>π</mi> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>×</mo> <mi>π</mi> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> </span> </p><p><span contenteditable="false"> <math> <mrow> <mi>i</mi> <mo>=</mo> <mfrac> <mrow> <mi>l</mi> <mtext> </mtext> <mtext> </mtext> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> </span>              </p><div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1757388464phpEF7clx.jpeg" alt="" width="97" height="76" loading="lazy"></picture> </div></div><p>Now, consider Amperian loop of radius small ‘r’ ln Amperian loop magnetic field will be tangential to the amperian loop.</p><p><span contenteditable="false"> <math> <mrow> <mstyle displaystyle="true"> <mrow> <mo>?</mo> <mrow> <mover accent="true"> <mi>B</mi> <mo>→</mo> </mover> <mo>.</mo> <mi>d</mi> <mover accent="true"> <mi>i</mi> <mo>→</mo> </mover> <mo>=</mo> <msub> <mrow> <mi>μ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mi>e</mi> <mi>n</mi> <mi>c</mi> <mi>l</mi> <mi>o</mi> <mi>s</mi> <mi>e</mi> <mi>d</mi> </mrow> </msub> </mrow> </mrow> </mstyle> </mrow> </math> </span>          (Ampere circuital law)</p><p><span contenteditable="false"> <math> <mrow> <mi>B</mi> <mo>=</mo> <mfrac> <mrow> <msub> <mrow> <mi>μ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mn>2</mn> <mi>π</mi> </mrow> </mfrac> <mfrac> <mrow> <mi>l</mi> </mrow> <mrow> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mi>r</mi> </mrow> </math> </span></p><p><span contenteditable="false"> <math> <mrow> <mi>B</mi> <mo>∝</mo> <mi>r</mi> </mrow> </math> </span> </p><p>              <img src="file:///C:/Users/RAJ~1.PAN/AppData/Local/Temp/msohtmlclip1/01/clip_image014.png" width="32" height="14"> </p>

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AB µ r² long straight wire with a circular cross-section having radius R, is carrying a steady current I: The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be:

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AB µ r2 long straight wire with a circular cross-section having radius R, is carrying a steady current I: The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be:

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AB µ r² long straight wire with a circular cross-section having radius R, is carrying a steady current I: The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be: