Electromagnetic Induction

JEE Main level, single correct, medium

39 of 153

Question 39 of 153

A conducting circular loop made of a thin wire has area 3.5×10−2m23.5\times {10}^{-2}{m}^{2} and resistance 10Ω10\Omega It is placed perpendicular to a time-dependent magnetic field B(t)=(0.4T)sin⁡⁡(50πt)B\left(t\right)=\left(0.4T\right)\sin ⁡\left(50\pi t\right) The field is uniform in space. Then the net charge flowing through the loop during t=0st=0s and t=10mst=10ms is close to

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Answer: (A) 1.4mC1.4mC

It is given that, the magnetic field,
B=0.4sin(50πt)B=0.4 sin\left(50\pi t\right)
and the area of the given circular conducting loop,
A=3.5×10−2m2A=3.5\times {10}^{-2}{m}^{2}
Magnetic flux,
ϕ=BA...(1)\phi =BA...(1)
Then, induced emf,

ϵ=∣dϕdt∣...(2)\epsilon =\left|\dfrac{d\phi }{dt}\right|...(2)
When resistance in the circuit is RR, the differential charge accumulated, based on current ii,

dQ=idt=ϵRdt=1Rdϕ⇒∆Q=1R∫dϕ=1R∆ϕdQ=idt=\dfrac{\epsilon }{R}dt=\dfrac{1}{R}d\phi \Rightarrow ∆Q=\dfrac{1}{R}\int d\phi =\dfrac{1}{R}∆\phi
Substituting given values and using equation (1)(1) and (2)(2),
∆Q=110[0.4sin⁡(50π×10×10−3)−0]×3.5×10−2⇒∆Q=0.4×3.5×10−210=1.4×10−3C∆Q=\dfrac{1}{10}\left[0.4\sin \left(50\pi \times 10\times {10}^{-3}\right)-0\right]\times 3.5\times {10}^{-2} \Rightarrow ∆Q=\dfrac{0.4\times 3.5\times {10}^{-2}}{10}=1.4\times {10}^{-3}C
⇒∆Q=1.4mC\Rightarrow ∆Q=1.4mC

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