We are asked to find the equivalent resistance R_PQ between points P and Q for a specific circuit configuration.

The circuit involves resistors arranged as follows:

R_PQ = (4 + 4)∥(4 + 4)

Step 1: Calculate Individual Branch Resistances

Each branch contains two 4-ohm resistors in series.

Therefore, the total resistance in each branch is:

R_branch = 4 + 4 = 8Ω

Step 2: Parallel Combination of Two Branches

Now, these two branches are in parallel.

The formula for the equivalent resistance R_parallel for two resistances R₁ and R₂ in parallel is:

${\mathrm{R}}_{\text{parallel }}=\frac{{\mathrm{R}}_1\times {\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}$ 

Substitute R₁ = 8Ω and R₂ = 8Ω :

${\mathrm{R}}_{\mathrm{PQ}}=\frac{8\times 8}{8+8}=\frac{64}{16}=4\mathrm{\Omega }$ 

Thus, the equivalent resistance R_PQ is 4Ω.

Now, the equivalent circuit has a resistance of R_eq = 4Ω.

We are given: 

- Magnetic field B = 5T,

- Side length of the square loop ℓ = 20cm = 0.20m

 - Steady current I = 2mA = 2 × 10⁻³A.

The induced emf e in the loop is given by :

e = Bv₀ℓ

The induced current I is given by:

$\mathrm{I}=\frac{\mathrm{e}}{{\mathrm{R}}_{\text{eq }}}$ 

Substitute e = Bv₀ℓ into the current equation:

$\mathrm{I}=\frac{{\mathrm{Bv}}_0\mathrm{\ell }}{{\mathrm{R}}_{\mathrm{eq}}}$ 

Now, substitute the known values into the equation: $2\times {10}^{-3}=\frac{5\times {\mathrm{v}}_0\times 0.2}{4}$ 

Simplify the equation:

$2\times {10}^{-3}=\frac{{\mathrm{v}}_0}{4}\times 1$ 

Multiply both sides by 4:

v₀ = 8 × 10⁻³m/s = 1cm/s

Final Answer :

The value of v₀ is 1cm/s, so that a steady current of 2mA flows in the loop.