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JEE Main 20199 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Evening Shift
Subject
Physics
Chapter
Magnetic Properties of Matter

JEE Main 2019 · 9 April · Evening Shift · Physics · Magnetic Properties of Matter

Two coils 'P' and 'Q' are separated by some distance. When a...

Two coils ′P′'P' and ′Q′'Q' are separated by some distance. When a current of 3A3A flows through coil ′P′,'{P}^{'}, a magnetic flux of 10−3Wb{10}^{- 3}Wb passes through ′Q′.'{Q}^{'} . No current is passed through ′Q′.'{Q}^{'} . When no current passes through ′P′'P' and a current of 2A2A passes through ′Q′,'{Q}^{'}, the flux through ′P′'P' is:

Options

  1. A3.67×10−3Wb3.67 \times {10}^{- 3}Wb
  2. B6.67×10−4Wb6.67 \times {10}^{- 4}Wb
  3. C3.67×10−4Wb3.67 \times {10}^{- 4}Wb
  4. D6.67×10−3Wb6.67 \times {10}^{- 3}Wb

Correct answer

B. 6.67×10−4Wb6.67 \times {10}^{- 4}Wb

Step-by-step solution

Since the coils are co-axial, the flux through the coils can be written as the product of their surface area and the magnetic field produced by the coil along its axis.

Consider the flux passing through Q

ϕQ=\mu0i1R22(R2+x2)32×\piR2=10−3Wb{ϕ}_{Q} = \dfrac{{\text{\mu}}_{0}{i}_{1}{R}^{2}}{2\left({R}^{2} + {x}^{2}\right)\dfrac{3}{2}} \times \text{\pi}{R}^{2} = {10}^{- 3}Wb

And the flux passing through P:
ϕP=\mu0i1r22(R2+x2)32×\piR2{ϕ}_{P} = \dfrac{{\text{\mu}}_{0}{i}_{1}{r}^{2}}{2\left({R}^{2} + {x}^{2}\right)\dfrac{3}{2}} \times \text{\pi}{R}^{2}

Taking the ratio
ϕPϕQ=i2i1×(R2+x2)32(R2+x2)32⇒ϕP10−3=i2i1\dfrac{{ϕ}_{P}}{{ϕ}_{Q}} = \dfrac{{i}_{2}}{{i}_{1}} \times \dfrac{{\left({R}^{2} + {x}^{2}\right)}^{\dfrac{3}{2}}}{{\left({R}^{2} + {x}^{2}\right)}^{\dfrac{3}{2}}} ⇒ \dfrac{{ϕ}_{P}}{{10}^{- 3}} = \dfrac{{i}_{2}}{{i}_{1}}
⇒ϕP10−3=23⇒ \dfrac{{ϕ}_{P}}{{10}^{- 3}} = \dfrac{2}{3}
∴ϕP=6.67×10−4Wb∴ {ϕ}_{P} = 6.67 \times {10}^{- 4}Wb

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