JEE Main 1-Sep-2021 Shift 2 Solved Paper

According to given circuit diagram, equivalent resistance between point P and Q.
RPQ=(4+4)||(4+4)
=

8×8

8+8

=4Ω
The equivalent circuit can be drawn as, Equivalent resistance, Req=4+1=5Ω
Magnetic field, B = 5T
The side of the square loop, l = 20 cm = 0.20 m
The steady value of the current, I=2‌mA=2×10−3A
Induced emf, e=Bv0l
Induced current, I=

e

Req

Substituting the values in the above equation, we get
2×10−3=

5×v0×0.2

5

⇒ v0=10−2m∕s=1‌cm∕s
∴ The value of v0=1‌cm∕ so that a steady current of 2‌mA flows in the loop.