Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.
This is a long answer type question as classified in NCERT Exemplar
The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vB d.
Since, switch S is closed at time t= 0. capacitor is charged by this potential difference. Let
Q ( t) is charge on the capacitor and current flows from A to B.
Now, the induced current I= -
On rearranging =+ dQ/dt =
On solving differential equation we get I= e-t/RC
<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vB d.</p><p>Since, switch S is closed at time t= 0. capacitor is charged by this potential difference. Let</p><p>Q ( t) is charge on the capacitor and current flows from A to B.</p><p>Now, the induced current I= <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>v</mi> <mi>B</mi> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math> </span> -<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>Q</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> <mi>C</mi> </mrow> </mrow> </mfrac> </math> </span></p><p>On rearranging =<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>Q</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> <mi>C</mi> </mrow> </mrow> </mfrac> </math> </span>+ dQ/dt =<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>v</mi> <mi>B</mi> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math> </span></p><p>On solving differential equation we get I=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>v</mi> <mi>B</mi> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math> </span> e<sup>-t/RC</sup></p>
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