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Electromagnetic Induction — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question
The rectangular wire-frame, shown in figure, has a width d, mass m, resistance R and a large length. A uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at t = 0.
  1. Find the acceleration of the frame when its speed has increased to v.
  2. Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find this velocity $v_0.$
  3. Show that the velocity at time t is given by $\text{v}=\text{v}_0\Big(1-\text{e}^{-\frac{\text{ft}}{\text{mv}_0}}\Big)$
✓

Answer

  1. emf developed = Bdv (when it attains a speed v)
Current $=\frac{\text{Bdv}}{\text{R}}$

Force $=\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}$

This force opposes the given force

Net F $=\text{F}-\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}=\text{RF}-\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}$

Net acceleration $=\frac{\text{RF}-\text{B}^2\text{d}^2\text{v}}{\text{mR}}$
  1. Velocity becomes constant when acceleration is 0.
$\frac{\text{F}}{\text{m}}-\frac{\text{B}^2\text{d}^2\text{v}_0}{\text{mR}}=0$

$\Rightarrow\frac{\text{F}}{\text{m}}=\frac{\text{B}^2\text{d}^2\text{v}_0}{\text{mR}}$

$\Rightarrow\text{v}_0=\frac{\text{FR}}{\text{B}^2\text{d}^2}$
  1. Velocity at line t
$\text{a}=-\frac{\text{dv}}{\text{dt}}$

$\Rightarrow\int_{0}^{\text{v}}\frac{\text{dv}}{\text{RF}-\text{l}^2\text{B}^2\text{v}}=\int_{0}^{\text{t}}\frac{\text{dt}}{\text{mR}}$

$\Rightarrow\Big[\text{l}_\text{n}\big[\text{RF}-\text{l}^2\text{B}^2\text{v}\big]\frac{1}{-\text{l}^2\text{B}^2}\Big]_0^{\text{v}} \ \Big[\frac{\text{t}}{\text{Rm}}\Big]_0^{\text{t}} $

$\Rightarrow\Big[\text{l}_\text{n}\big(\text{RF}-\text{l}^2\text{B}^2\text{v}\big)\Big]_{0}^{\text{v}}=\frac{-\text{tl}^2\text{B}^2}{\text{Rm}}$

$\Rightarrow\text{l}_\text{n}\big(\text{RF}-\text{l}^2\text{B}^2\text{v}\big)-\text{ln}(\text{RF})=\frac{-\text{t}^2\text{B}^2\text{t}}{\text{Rm}}$

$\Rightarrow1-\frac{\text{l}^2\text{B}^2\text{v}}{\text{Rf}}=\text{e}^{\frac {-\text{l}^2\text{B}^2\text{t}}{\text{Rm}}}$

$\Rightarrow\frac{\text{l}^2\text{B}^2\text{v}}{\text{Rf}}=1-\text{e}^{\frac {-\text{l}^2\text{B}^2\text{t}}{\text{Rm}}}$

$\Rightarrow\text{v}=\frac{\text{FR}}{\text{l}^2\text{B}^2}\Big(1-\text{e}^{\frac{-\text{l}^2\text{B}^2\text{v}_0\text{t}}{\text{Rv}_0\text{m}}}\Big)=\text{v}_0(1-\text{e}^{-\text{Fv}_0\text{m}})$

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