Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the line integral ζ ( L ) = | ${\int }_{-\mathit{L}}^{\mathit{L}}\mathit{B}.\mathit{d}\mathit{l}$ | taken along z-axis.

(a) Show that ? ( ) monotonically increases with L.

(b) Use an appropriate Amperian loop to show that ζ ( ∞ ) = ?0? , where I is the current in the wire.

(c) Verify directly the above result.

(d) Suppose we replace the circular coil by a square coil of sides R carrying the same current I. What can you say about ζ ( L ) and ζ ( ∞ ) ?

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1

(b) $\int B.dl={\mu }_{o}I$ but when L $\to$ $\infty$

So B $\propto$ 1/r³

(c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.

 

B= $\frac{{\mu }_{0}I{R}^2}{2{(Z^2+R^2)}^{3/2}}$

Z=Rtan $\theta$

dz=Rsec² $\theta d\theta$

${\int }_{-\infty }^{\infty }{B}_{z}dz=\frac{{\mu }_{0}I}{2}{\int }_{-\pi /2}^{\pi /2}cos\theta d\theta$ = ${\mu }_{0}I$

<p>This is a Long Answer Type Questions as classified in NCERT Exemplar</p><p><strong>Explanation-&nbsp;</strong>B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1</p><p> (b)&nbsp;<math><mo>∫</mo><mrow><mrow><mi>B</mi><mo>.</mo><mi>d</mi><mi>l</mi><mo>=</mo><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mi>o</mi></mrow></mrow></msub><mi>I</mi></mrow></mrow></math>&nbsp;but when L&nbsp;<math><mo>→</mo></math>&nbsp;<math><mi>∞</mi></math></p><p>So B&nbsp;<math><mo>∝</mo></math> 1/r³</p><p> (c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1742970473phpxObRU4_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1742970473phpxObRU4.jpeg" alt="" width="178" height="117"></picture></div></div><p>&nbsp;</p><p>B=&nbsp;<math><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>2</mn><msup><mrow><mrow><mo> (</mo><msup><mrow><mrow><mi>Z</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>)</mo></mrow></mrow><mrow><mrow><mn>3</mn><mo>/</mo><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></p><p>Z=Rtan&nbsp;<math><mi>θ</mi></math></p><p>dz=Rsec² <math><mi>θ</mi><mi>d</mi><mi>θ</mi></math></p><p><math><msubsup><mo>∫</mo><mrow><mo>-</mo><mi>∞</mi></mrow><mrow><mi>∞</mi></mrow></msubsup><mrow><mrow><msub><mrow><mrow><mi>B</mi></mrow></mrow><mrow><mrow><mi>z</mi></mrow></mrow></msub><mi>d</mi><mi>z</mi></mrow></mrow><mo>=</mo><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><msubsup><mo>∫</mo><mrow><mo>-</mo><mi>π</mi><mo>/</mo><mn>2</mn></mrow><mrow><mi>π</mi><mo>/</mo><mn>2</mn></mrow></msubsup><mrow><mrow><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mi>d</mi><mi>θ</mi></mrow></mrow></math>&nbsp;=&nbsp;<math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi></math></p>

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