1. Identify the effect involved

This is the Hall effect. A current-carrying metallic strip is placed in a magnetic field.

• Current is along $+x$.
• Magnetic field is along $+y$.
• Charge carriers are electrons.
• Hence electrons experience magnetic deflection along the $z$-direction, producing a Hall electric field along $z$.

The Hall voltage is the potential difference between the two opposite faces parallel to the $x$-$y$ plane.

2. Drift velocity of electrons

If $n$ is the number of free electrons per unit volume, then

$I = n e A v_d$

where cross-sectional area perpendicular to current is

$A = wd$

So,

$v_d = \frac{I}{n e w d}$

3. Balance of magnetic and electric forces

At equilibrium,

$eE_H = e v_d B$

Therefore,

$E_H = v_d B$

Substitute $v_d$:

$E_H = \frac{I B}{n e w d}$

4. Hall voltage across thickness direction

The potential difference between the two faces separated along $z$ by distance $d$ is

$V = E_H d$

Thus,

$V = \frac{I B}{n e w d} \cdot d = \frac{I B}{n e w}$

So the Hall voltage is:

$\boxed{V \propto \frac{1}{w}}$

and is independent of thickness $d$.

5. Check each option

Since both strips are of the same material and carry the same current $I$ in the same magnetic field $B$,

$V = \frac{IB}{ne w}$

Hence only width matters.

Option A

If $w_1 = w_2$ and $d_1 = 2d_2$,

since voltage is independent of $d$,

$V_1 = V_2$

So statement $V_2 = 2V_1$ is false.

Option B

If $w_1 = w_2$ and $d_1 = 2d_2$,

$V_2 = V_1$

So B is correct.

Option C

If $w_1 = 2w_2$ and $d_1 = d_2$,

$V_1 = \frac{IB}{ne w_1}, \qquad V_2 = \frac{IB}{ne w_2}$

Since $w_1 = 2w_2$,

$V_1 = \frac{IB}{ne (2w_2)} = \frac{1}{2} \frac{IB}{ne w_2} = \frac{V_2}{2}$

Thus,

$V_2 = 2V_1$

So C is correct.

Option D

From above,

$V_2 = 2V_1 \neq V_1$

So D is false.

6. Derived answer

The correct options are

$\boxed{\text{B and C}}$

7. Comparison with stored answer

Stored answer: A, D

Our derived answer: B, C

These do not match.

The likely reason is that Hall voltage across the faces separated by thickness $d$ is

$V = E_H d = \left(\frac{IB}{ne w d}\right)d = \frac{IB}{ne w}$

so thickness cancels out, and the voltage depends only on width $w$ for fixed $I$, $B$, and material.