6.16 (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

6.16 Let us take a small element dy in the loop, at a distance y from the long straight wire.

 

Magnetic flux associated with element dy, d $\varnothing$ = BdA, where

dA = Area of the element dy = a dy

Magnetic flux at distance y, B = $\frac{{\mu }_{0}I}{2\pi y}$ , where

I = current in the wire

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Therefore,

d $\varnothing$ = $\frac{{\mu }_{0}I}{2\pi y}$ a dy = $\frac{{\mu }_{0}Ia}{2\pi }$ $\frac{dy}{y}$

$\varnothing =\frac{{\mu }_{0}Ia}{2\pi }\int \frac{dy}{y}$

Now from the figure, the range of y is x to x+a. Hence,

$\varnothing =\frac{{\mu }_{0}Ia}{2\pi }{\int }_x^{x+a}\frac{dy}{y}$ = $\frac{{\mu }_{0}Ia}{2\pi }{\left[{\log }_e?y\right]}_x^{a+x}$ = $\frac{{\mu }_{0}Ia}{2\pi }{\log }_e?(\frac{a+x}{x})$

For mutual inductance M, the flux is given as

$\varnothing =MI$ . Hence

$MI=\frac{{\mu }_{0}Ia}{2\pi }{\log }_e?(\frac{a+x}{x})$

M = $\frac{{\mu }_{0}a}{2\pi }{\log }_e?(\frac{a}{x}+1)$

Emf induced in the loop, e = Bav

= ( $\frac{{\mu }_{0}I}{2\pi x}$ ) $\times$ av

For I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s

...more

6.16 Let us take a small element dy in the loop, at a distance y from the long straight wire.

 

Magnetic flux associated with element dy, d $\varnothing$ = BdA, where

dA = Area of the element dy = a dy

Magnetic flux at distance y, B = $\frac{{\mu }_{0}I}{2\pi y}$ , where

I = current in the wire

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Therefore,

d $\varnothing$ = $\frac{{\mu }_{0}I}{2\pi y}$ a dy = $\frac{{\mu }_{0}Ia}{2\pi }$ $\frac{dy}{y}$

$\varnothing =\frac{{\mu }_{0}Ia}{2\pi }\int \frac{dy}{y}$

Now from the figure, the range of y is x to x+a. Hence,

$\varnothing =\frac{{\mu }_{0}Ia}{2\pi }{\int }_x^{x+a}\frac{dy}{y}$ = $\frac{{\mu }_{0}Ia}{2\pi }{\left[{\log }_e?y\right]}_x^{a+x}$ = $\frac{{\mu }_{0}Ia}{2\pi }{\log }_e?(\frac{a+x}{x})$

For mutual inductance M, the flux is given as

$\varnothing =MI$ . Hence

$MI=\frac{{\mu }_{0}Ia}{2\pi }{\log }_e?(\frac{a+x}{x})$

M = $\frac{{\mu }_{0}a}{2\pi }{\log }_e?(\frac{a}{x}+1)$

Emf induced in the loop, e = Bav

= ( $\frac{{\mu }_{0}I}{2\pi x}$ ) $\times$ av

For I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s

e = $\frac{4\pi \times {10}^{-7}\times 50\times 0.1\times 10}{2\pi \times 0.2}$ = 5 $\times {10}^{-5}$ V

less

<p><strong>6.16 </strong>Let us take a small element dy in the loop, at a distance y from the long straight wire.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728554591phpadUKXi_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728554591phpadUKXi.jpeg" alt="" width="366" height="271"></picture></div></div><p>&nbsp;</p><p>Magnetic flux associated with element dy, d&nbsp;<span title="Click to copy mathml"><math><mi></mi><mi>∅</mi></math></span>&nbsp;= BdA, where</p><p>dA = Area of the element dy = a dy</p><p>Magnetic flux at distance y, B =&nbsp;<span title="Click to copy mathml"><math><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><mi>π</mi><mi>y</mi></mrow></mrow></mfrac></math></span>&nbsp;, where</p><p>I = current in the wire</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math></span>&nbsp;= Permeability of free space = 4&nbsp;<span title="Click to copy mathml"><math><mi>π</mi><mi></mi><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup></math></span>&nbsp;T m&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>A</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math></span></p><p>Therefore,</p><p>d&nbsp;<span title="Click to copy mathml"><math><mi></mi><mi>∅</mi></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><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><mi>π</mi><mi>y</mi></mrow></mrow></mfrac></math></span>&nbsp;a dy =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow></mfrac></math></span>&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>d</mi><mi>y</mi></mrow></mrow><mrow><mrow><mi>y</mi></mrow></mrow></mfrac></math></span></p><p><span title="Click to copy mathml"><math><mi>∅</mi><mo>=</mo><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow></mfrac><mo>∫</mo><mrow><mrow><mfrac><mrow><mrow><mi>d</mi><mi>y</mi></mrow></mrow><mrow><mrow><mi>y</mi></mrow></mrow></mfrac></mrow></mrow></math></span></p><p>Now from the figure, the range of y is x to x+a. Hence,</p><p><span title="Click to copy mathml"><math><mi>∅</mi><mo>=</mo><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow></mfrac><msubsup><mo>∫</mo><mrow><mi>x</mi></mrow><mrow><mi>x</mi><mo>+</mo><mi>a</mi></mrow></msubsup><mrow><mrow><mfrac><mrow><mrow><mi>d</mi><mi>y</mi></mrow></mrow><mrow><mrow><mi>y</mi></mrow></mrow></mfrac></mrow></mrow></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow></mfrac><msubsup><mrow><mrow><mfenced open="[" close="]" separators="|"><mrow><mrow><mrow><mrow><msub><mrow><mrow><mi mathvariant="normal">log</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></mrow><mo>?</mo><mrow><mi>y</mi></mrow></mrow></mrow></mrow></mfenced></mrow></mrow><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mi>a</mi><mo>+</mo><mi>x</mi></mrow></mrow></msubsup></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow></mfrac><mi></mi><mrow><mrow><msub><mrow><mrow><mi mathvariant="normal">log</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></mrow><mo>?</mo><mrow><mo>(</mo><mfrac><mrow><mrow><mi>a</mi><mo>+</mo><mi>x</mi></mrow></mrow><mrow><mrow><mi>x</mi></mrow></mrow></mfrac></mrow></mrow><mo>)</mo></math></span></p><p>For mutual inductance M, the flux is given as</p><p><span title="Click to copy mathml"><math><mi>∅</mi><mo>=</mo><mi>M</mi><mi>I</mi></math></span>&nbsp;. Hence</p><p><span title="Click to copy mathml"><math><mi>M</mi><mi>I</mi><mo>=</mo><mi></mi><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>I</mi><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow></mfrac><mi></mi><mrow><mrow><msub><mrow><mrow><mi mathvariant="normal">log</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></mrow><mo>?</mo><mrow><mo>(</mo><mfrac><mrow><mrow><mi>a</mi><mo>+</mo><mi>x</mi></mrow></mrow><mrow><mrow><mi>x</mi></mrow></mrow></mfrac></mrow></mrow><mo>)</mo></math></span></p><p>M =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow></mfrac><mi></mi><mrow><mrow><msub><mrow><mrow><mi mathvariant="normal">log</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></mrow><mo>?</mo><mrow><mo>(</mo><mfrac><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>x</mi></mrow></mrow></mfrac><mo>+</mo><mi></mi><mn>1</mn><mo>)</mo></mrow></mrow></math></span></p><p>Emf induced in the loop, e = Bav</p><p>= (&nbsp;<span title="Click to copy mathml"><math><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><mi>π</mi><mi>x</mi></mrow></mrow></mfrac></math></span>&nbsp;)&nbsp;<span title="Click to copy mathml"><math><mo>×</mo></math></span>&nbsp;av</p><p>For I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s</p><p>e =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>4</mn><mi>π</mi><mi></mi><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup><mo>×</mo><mn>50</mn><mo>×</mo><mn>0.1</mn><mo>×</mo><mn>10</mn></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi><mo>×</mo><mn>0.2</mn><mi></mi></mrow></mrow></mfrac></math></span>&nbsp;= 5&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>5</mn><mi></mi></mrow></mrow></msup></math></span>&nbsp;V</p>

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