Verified Solution • 2025–26 Session moderate

QJEEM20Ph57f2: JEE Main 2020 Physics - Class 12 JEEM20Ph57f2 Solution: A uniform magnetic field B exists in a direction perpendicular…

A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dBdt{{dB} \over {dt}} = 0.032 Ts–1. The induced current in the loop is close to (Resistivity of the metal wire is 1.23 ×\times 10–8 Ω\Omega m)

Multiple Choice Options

  • A 0.53 A
  • B 0.43 A
  • C 0.34 A
  • D 0.61 A

Step-by-Step Verified Solution

We know, ϕ=BA\phi = BA

Also, E=dϕdt=AdBdtE = {{d\phi } \over {dt}} = {{AdB} \over {dt}}

E=l2dBdtE = {l^2}{{dB} \over {dt}}

i=ERi = {E \over R}

= l2dBdtρlA{{{l^2}{{dB} \over {dt}}} \over {{{\rho l} \over A}}}

=l2pldBdtA= {{{l^2}} \over {pl}}{{dB} \over {dt}}A

= l2πR2ρldBdt{{{{l^2}\pi {R^2}} \over {\rho l}}{{dB} \over {dt}}}

∴\therefore i=304×304×10−4×0.032×4×10−6×π1.23×10−8×30×10−2×103i = {{30} \over 4} \times {{30} \over 4} \times {{{{10}^{ - 4}} \times 0.032 \times 4 \times {{10}^{ - 6}} \times \pi } \over {1.23 \times {{10}^{ - 8}} \times 30 \times {{10}^{ - 2}} \times {{10}^3}}}

⇒\Rightarrow i=240×π×10−101.23×10−7i = {{240 \times \pi \times {{10}^{ - 10}}} \over {1.23 \times {{10}^{ - 7}}}}

⇒\Rightarrow i=240×3.14×10−31.23i = {{240 \times 3.14 \times {{10}^{ - 3}}} \over {1.23}}

=753.61.23×10−3= {{753.6} \over {1.23}} \times {10^{ - 3}}

⇒\Rightarrow i=612.68×10−3=0.61Ai = 612.68 \times {10^{ - 3}} = 0.61A

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