A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic- field. If z is the vertical direction, the z-component of magnetic field is Bz=B0(1+ z). If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, and acceleration due to gravity g.

This is a long answer type question as classified in NCERT Exemplar ? = B z π l 2 = B 0 ( 1 + λ z ) ( π l 2 ) According to faraday’s law e= - d ∅ d t and also ohm’s law V=IRSo by equating these twoWe get Bo( π l 2 ) λ d z d t = I R I= π l 2 B o λ v R And energy lost/second = I2R= ( π l 2 λ ) 2 B o 2 v 2 R As change in PE=mg d z d t = m g v but if we compare both energy thenMgv= ( π l 2 λ ) 2 B o 2 v 2 R or v=

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

A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic- field. If z is the vertical direction, the z-component of magnetic field is B_z=B₀(1+ z). If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, and acceleration due to gravity g.

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Payal Gupta | Contributor-Level 10

4 months ago

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

$? = B_{z} (π l^{2}) = B_{0} (1 + λ z) (π l^{2})$

According to faraday’s law e= - $\frac{d \emptyset}{d t}$ and also ohm’s law V=IR

So by equating these two

We get B_o( $π l^{2}$ ) $λ \frac{d z}{d t} = I R$

I= $\frac{π l^{2} B_{o} λ v}{R}$

And energy lost/second = I²R= $\frac{{(π l^{2} λ)}^{2} B_{o}^{2} v^{2}}{R}$

As change in PE=mg $\frac{d z}{d t} = m g v$ but if we compare both energy then

Mgv= $\frac{{(π l^{2} λ)}^{2} B_{o}^{2} v^{2}}{R}$ or v=

This is a long answer type question as classified in NCERT Exemplar <math> <mi>?</mi> <mo>=</mo> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mi>z</mi> </mrow> </mrow> </msub> <mfenced separators="|"> <mrow> <mrow> <mi>π</mi> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> <mo>=</mo> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>λ</mi> <mi>z</mi> <mo>)</mo> <mo>(</mo> <mi>π</mi> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>)</mo> </math> According to faraday’s law e= - <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>∅</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> and also ohm’s law V=IRSo by equating these twoWe get Bo( <math> <mi>π</mi> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> ) <math> <mi>λ</mi> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>z</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mi>I</mi> <mi>R</mi> </math> I= <math> <mfrac> <mrow> <mrow> <mi>π</mi> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> <mi>λ</mi> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math> And energy lost/second = I2R= <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mi>π</mi> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>λ</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msubsup> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math> As change in PE=mg <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>z</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mi>m</mi> <mi>g</mi> <mi>v</mi> </math> but if we compare both energy thenMgv= <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mi>π</mi> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>λ</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msubsup> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math>  or v= <img src="https://images.shiksha.com/mediadata/images/articles/1743064615phpu9KoUP.jpeg" alt="" width="51" height="37">

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