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.