A cell having N turns is would tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil:

Option 1 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>μ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> <mi>I</mi> </mrow> <mrow> <mn>8</mn> </mrow> </mfrac> <mrow> <mo>[</mo> <mrow> <mfrac> <mrow> <mi>a</mi> <mo>+</mo> <mi>b</mi> </mrow> <mrow> <mi>a</mi> <mo>−</mo> <mi>b</mi> </mrow> </mfrac> </mrow> <mo>]</mo> </mrow> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>μ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> <mi>I</mi> <mi>N</mi> </mrow> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mrow> <mi>b</mi> <mo>−</mo> <mi>a</mi> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mi>l</mi> <mi>o</mi> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mi>e</mi> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mi>a</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>μ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> <mi>I</mi> </mrow> <mrow> <mn>4</mn> <mrow> <mo>(</mo> <mrow> <mi>a</mi> <mo>−</mo> <mi>b</mi> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mrow> <mo>[</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>a</mi> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>b</mi> </mrow> </mfrac> </mrow> <mo>]</mo> </mrow> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>μ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> <mi>I</mi> </mrow> <mrow> <mn>8</mn> </mrow> </mfrac> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>a</mi> <mo>−</mo> <mi>b</mi> </mrow> <mrow> <mi>a</mi> <mo>+</mo> <mi>b</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>

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Vishal Baghel | Contributor-Level 10

4 months ago

Correct Option - 2

Detailed Solution:

Let us consider an elementary ring of radius r and thickness dr in which current is flowing.

So, No. of turns in this elementary ring

$d N = (\frac{N}{b - a}) d r$

$∴ {(d B)}_{a t c e n t r e} = \frac{μ_{0} l d N}{2 r}$

$B = \frac{μ_{0} l N}{2 (b - a)} l n \frac{b}{a}$

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Physics Moving Charges and Magnetism 2025

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