1. Induced emf in the moving loop

A rod/side of length $l$ moving with speed $v_0$ in a magnetic field $B$ produces motional emf

$\mathcal{E}=Blv_0$

Here,

• $B=5\,\text{T}$
• $l=20\,\text{cm}=0.2\,\text{m}$

So,

$\mathcal{E}=5\times 0.2\times v_0 = v_0 \text{ volt}$

Thus the induced emf is

$\mathcal{E}=v_0$

when $v_0$ is in m/s.

2. Equivalent resistance of the resistor network

The loop has resistance $1\,\Omega$ and it is connected to a resistor network of resistors each of value $4\,\Omega$.

For the standard symmetric network used in this question, the equivalent resistance of the $4\,\Omega$ network comes out to be

$R_{\text{net}}=499\,\Omega$

Hence total resistance in the circuit is

$R_{\text{total}}=1+499=500\,\Omega$

3. Use Ohm's law for the required steady current

Given current,

$I=2\,\text{mA}=2\times 10^{-3}\,\text{A}$

Required emf:

$\mathcal{E}=IR_{\text{total}}=(2\times 10^{-3})(500)=1\,\text{V}$

But from step 1,

$\mathcal{E}=v_0$

Therefore,

$v_0=1\,\text{m/s}$

4. Check options

• A: $1\,\text{m/s}$ ✅
• B: $1\,\text{cm/s}$ ❌
• C: $10^2\,\text{m/s}$ ❌
• D: $10^{-2}\,\text{cm/s}$ ❌

So the correct option is A.