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Electromagnetic Induction Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Six

A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires as shown in figure. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field B = B(t) k

(i) Write down an equation for the acceleration of the wire XY

(ii) If B is independent of time, obtain v(t), assuming v(0) = u₀

(iii) For (ii), show that the decrease in kinetic energy of XY equals the heat lost in

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Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

Let us assume that the parallel wires at are y= 0, i.e., along x-axis and y= l. At t= 0, XY has x = 0 i.e., along y-axis.

(i) Let the wire be at x x = (t) at time t.

The magnetic flux linked with the loop is given by

φ = B.A= Bacos0= BA

magnetic flux = B(t)(lx(t))

total emf = emf due to change in field along XYAC+ the motional emf across XY

E= - $\frac{d \emptyset}{d t}$ = - $\frac{d B (t)}{d t}$ lx(t)- B(t)lv(t)

Electric current in clockwise direction is I= E/r

So force is F= ilBsin90=ilB

Force = $\frac{I B (t)}{R}$ [- $\frac{d B (t)}{d t} I x (t) - B (t) I v (t)$ ]i

Applying newton 2^nd law

m $\frac{d^{2} x}{d t^{2}}$ = $\frac{I^{2} B (t) d B}{R d t} x (t)$ - $\frac{I^{2} B^{2} (t) d x}{R d t}$

(ii) $\frac{d B}{d t}$ =0

Substituting in eqn 1

$\frac{d^{2} x}{d t^{2}}$ + $\frac{I^{2} B^{2} (t) d x}{m R d t}$ =0

$\frac{d v}{d t} + \frac{I^{2} B^{2} v}{m R}$ =o

V= Aexp $(\frac{I^{2} B^{2} v}{m R}$ )=0

(iii) p= I²R= $\frac{I^{2} B^{2} v^{2} R}{R^{2}} = \frac{B^{2} I^{2}}{R} u_{0}^{2}$ exp

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This is a long answer type question as classified in NCERT Exemplar

Let us assume that the parallel wires at are y= 0, i.e., along x-axis and y= l. At t= 0, XY has x = 0 i.e., along y-axis.

(i) Let the wire be at x x = (t) at time t.

The magnetic flux linked with the loop is given by

φ = B.A= Bacos0= BA

magnetic flux = B(t)(lx(t))

total emf = emf due to change in field along XYAC+ the motional emf across XY

E= - $\frac{d \emptyset}{d t}$ = - $\frac{d B (t)}{d t}$ lx(t)- B(t)lv(t)

Electric current in clockwise direction is I= E/r

So force is F= ilBsin90=ilB

Force = $\frac{I B (t)}{R}$ [- $\frac{d B (t)}{d t} I x (t) - B (t) I v (t)$ ]i

Applying newton 2^nd law

m $\frac{d^{2} x}{d t^{2}}$ = $\frac{I^{2} B (t) d B}{R d t} x (t)$ - $\frac{I^{2} B^{2} (t) d x}{R d t}$

(ii) $\frac{d B}{d t}$ =0

Substituting in eqn 1

$\frac{d^{2} x}{d t^{2}}$ + $\frac{I^{2} B^{2} (t) d x}{m R d t}$ =0

$\frac{d v}{d t} + \frac{I^{2} B^{2} v}{m R}$ =o

V= Aexp $(\frac{I^{2} B^{2} v}{m R}$ )=0

(iii) p= I²R= $\frac{I^{2} B^{2} v^{2} R}{R^{2}} = \frac{B^{2} I^{2}}{R} u_{0}^{2}$ exp(-2I²B²tlmR)

This proves decrease in kinetic energy.

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This is a long answer type question as classified in NCERT ExemplarLet us assume that the parallel wires at are y= 0, i.e., along x-axis and y= l. At t= 0, XY has x = 0 i.e., along y-axis.(i) Let the wire be at x x = (t) at time t.The magnetic flux linked with the loop is given byφ = B.A= Bacos0= BAmagnetic flux = B(t)(lx(t))total emf = emf due to change in field along XYAC+ the motional emf across XYE= - <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>∅</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math>  = - <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>B</mi> <mo>(</mo> <mi>t</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> lx(t)- B(t)lv(t)Electric current in clockwise direction is I= E/rSo force is F= ilBsin90=ilBForce = <math> <mfrac> <mrow> <mrow> <mi>I</mi> <mi>B</mi> <mo>(</mo> <mi>t</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math> [- <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>B</mi> <mfenced separators="|"> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> <mi>I</mi> <mi>x</mi> <mfenced separators="|"> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> </mfenced> <mo>-</mo> <mi>B</mi> <mo>(</mo> <mi>t</mi> <mo>)</mo> <mi>I</mi> <mi>v</mi> <mo>(</mo> <mi>t</mi> <mo>)</mo> </math> ]iApplying newton 2nd lawm <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>B</mi> <mfenced separators="|"> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> </mfenced> <mi>d</mi> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> <mi>x</mi> <mo>(</mo> <mi>t</mi> <mo>)</mo> </math> - <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mfenced separators="|"> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> </mfenced> <mi>d</mi> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> (ii) <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> =0Substituting in eqn 1 <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> + <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mfenced separators="|"> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> </mfenced> <mi>d</mi> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> <mi>R</mi> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> =0 <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> <mi>R</mi> </mrow> </mrow> </mfrac> </math> =oV= Aexp <math> <mo>(</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> <mi>R</mi> </mrow> </mrow> </mfrac> </math> )=0(iii) p= I2R= <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> <msubsup> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </math> exp(-2I2B2tlmR)This proves decrease in kinetic energy.

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