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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Physics
Chapter
Electromagnetic Induction

BITSAT 2014 · Physics · Electromagnetic Induction

A coil 10 turns and a resistance of 20 is connected in series with...

A coil 10 turns and a resistance of 20Ω20 \Omega is connected in series with B.G of resistance 30Ω30 \Omega. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction 10−2T10^{-2} \mathrm{ T}. If it is now turned through an angle of 60∘60^{{}^{\circ}} about an axis in its plane. Find the charge induced in the coil. (Area of a coil =10−2m2=10^{-2} \mathrm{ m}^{2} )

Options

  1. A2×10−5C2 \times 10^{-5} \mathrm{C}
  2. B3.2×10−5C3.2 \times 10^{-5} \mathrm{C}
  3. C1×10−5C1 \times 10^{-5} \mathrm{C}
  4. D5.5×10−5C5.5 \times 10^{-5} \mathrm{C}

Correct answer

C. 1×10−5C1 \times 10^{-5} \mathrm{C}

Step-by-step solution

Given: n=10\mathrm{n}=10 turns, Rcoil =20Ω,RG=30Ω\mathrm{R}_{\text{coil }}=20 \Omega, \mathrm{R}_{\mathrm{G}}=30 \Omega, Total resistance in the circuit =20+30=50ΩA=10−2m2,B=10−2T,ϕ1=0∘,ϕ2=60∘q=ϕ1−ϕ2R=BnAcos⁡θ1−BnAcos⁡θ2R=Bn⁡A(cos⁡0−cos⁡60)R=BnA(1−0.5)R=0.5×10−2×10×10−250=50×10−550=1×10−5C=20+30=50 \Omega\mathrm{A}=10^{-2} \mathrm{ m}^{2}, \mathrm{ B}=10^{-2} \mathrm{ T}, \phi_{1}=0^{{}^{\circ}}, \phi_{2}=60^{{}^{\circ}}\mathrm{q}=\frac{\phi_{1}-\phi_{2}}{\mathrm{R}}=\frac{\mathrm{Bn} \mathrm{A} \cos \theta_{1}-\mathrm{Bn} \mathrm{A} \cos \theta_{2}}{\mathrm{R}}=\frac{\operatorname{Bn} \mathrm{A}(\cos 0-\cos 60)}{\mathrm{R}}=\frac{\mathrm{BnA}(1-0.5)}{\mathrm{R}}=\frac{0.5 \times 10^{-2} \times 10 \times 10^{-2}}{50}=\frac{50 \times 10^{-5}}{50}=1 \times 10^{-5} \mathrm{C} \quad (Charge induced in a coil)
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