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AIIMS 2017
Exam
AIIMS
Year
2017
Subject
Physics
Chapter
Electromagnetic Induction

AIIMS 2017 · Physics · Electromagnetic Induction

As shown in figure, two vertical conducting rails separated by...

As shown in figure, two vertical conducting rails separated by distance 1.0 m1.0 \mathrm{~m} are placed parallel to zz-axis. At z=0z=0, a capacitor of 0.15 F0.15 \mathrm{~F} is connected between the rails and a metal rod of mass 100gm100 \mathrm{gm} placed across the rails slides down along the rails. If a constant magnetic fields of 2.0 T2.0 \mathrm{~T} exists perpendicular to the plane of the rails, what is the acceleration of the rod? (Take g=9.8 m/s2g=9.8 \mathrm{~m} / \mathrm{s}^2 ) As shown in figure, two vertical conducting rails separated by distance 1.0 ~m are placed parallel to z -axis. At z=0 ,

Options

  1. A2.5 m/s22.5 \mathrm{~m} / \mathrm{s}^2
  2. B1.4 m/s21.4 \mathrm{~m} / \mathrm{s}^2
  3. C9.8 m/s29.8 \mathrm{~m} / \mathrm{s}^2
  4. D00

Correct answer

B. 1.4 m/s21.4 \mathrm{~m} / \mathrm{s}^2

Step-by-step solution

Due to motion of rod, emf induced across capacitor, ε=Blv∴\varepsilon=B l v \therefore \quad Charge stored in capacitor, Q=C(Blv)I=dQdt=CBldvdt=CBlaQ=C(B l v) I=\frac{d Q}{d t}=C B l \frac{d v}{d t}=C B l a Force opposing the downward motion, Fm=BIl∴Fm=B(CBla)l=B2l2CaF_m=B I l \therefore \quad F_m=B(C B l a) l=B^2 l^2 C a Net force on rod, Fnet =W−Fm=mg−B2l2Ca∴ma=mg−B2l2CaF_{\text{net }}=W-F_m=m g-B^2 l^2 C a \therefore \quad m a=m g-B^2 l^2 C a or a=mg(m+B2l2C)\quad a=\frac{m g}{\left(m+B^2 l^2 C\right)}So,So,\quad a=\frac{0.1 \times 9.8}{\left(0.1+2^2 \times 1^2 \times 0.15\right)}=1.4 \mathrm{ m} / $$\mathrm{s}^2
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