JEE Challenger
More from Moving Charges and Magnetism
PhyMedAdv 2015 • Paper 2

Hall Voltage Dependence on Carrier Density and Magnetic Field

Comprehension Passage

In a thin rectangular metallic conductor strip, a steady current II flows along the positive xx-axis, as shown in the illustration. The length, width, and thickness of this strip are denoted by ll, ww, and dd, respectively. A uniform external magnetic field B⃗\vec{B} acts along the positive yy-direction. Consequently, the mobile charge carriers experience a transverse magnetic deflection along the zz-direction. This causes accumulation of charge carriers on the planar surface PQRSPQRS and induces an equal and opposite charge layer on the surface opposite to PQRSPQRS. Therefore, a potential difference develops along the zz-direction. The accumulation of charges persists until the transverse magnetic force is counterbalanced by the electrostatic force. Assume that the electric current is distributed uniformly across the strip's cross-section and is conducted solely by electrons.

Passage Diagram 1

Consider two metallic strips (1 and 2) having identical geometric dimensions (length ll, width ww, and thickness dd), but with carrier number densities n1n_1 and n2n_2, respectively. Strip 1 is situated in a magnetic field B1B_1 and strip 2 is situated in a magnetic field B2B_2, both oriented along the positive yy-axis. Let V1V_1 and V2V_2 denote the potential differences produced between points KK and MM in strips 1 and 2, respectively. Assuming the current II passing through both strips is identical, which of the following option(s) is/are correct?

Options

A

If B1=B2B_1 = B_2 and n1=2n2n_1 = 2n_2, then V2=2V1V_2 = 2V_1

Correct
B

If B1=B2B_1 = B_2 and n1=2n2n_1 = 2n_2, then V2=V1V_2 = V_1

C

If B1=2B2B_1 = 2B_2 and n1=n2n_1 = n_2, then V2=0.5V1V_2 = 0.5V_1

Correct
D

If B1=2B2B_1 = 2B_2 and n1=n2n_1 = n_2, then V2=V1V_2 = V_1

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceElectric Field

Step-by-Step Solution

To find the relationship between the potential difference (Hall voltage) VV and the parameters of the metallic strip, let us analyze the forces acting on the charge carriers (electrons) at steady state.

1. Drift Velocity of Charge Carriers:

The current II flowing through the strip of cross-sectional area A=w×dA = w \times d along the xx-direction is given by: I=neAvd=newdvdI = n e A v_d = n e w d v_d

where:

  • nn is the charge carrier density,
  • ee is the elementary charge,
  • ww is the width of the strip along the zz-direction,
  • dd is the thickness of the strip along the yy-direction,
  • vdv_d is the drift velocity of the electrons.

Thus, the drift velocity is: vd=Inewdv_d = \frac{I}{n e w d}

2. Hall Electric Field and Potential Difference:

A uniform magnetic field B⃗=Bj^\vec{B} = B\hat{j} is applied along the yy-direction. The magnetic force on an electron moving with drift velocity v⃗d=−vdi^\vec{v}_d = -v_d\hat{i} is: F⃗B=−e(v⃗d×B⃗)=−e(−vdi^×Bj^)=evdBk^\vec{F}_B = -e (\vec{v}_d \times \vec{B}) = -e (-v_d\hat{i} \times B\hat{j}) = e v_d B \hat{k}

This causes accumulation of electrons along the zz-direction, establishing a transverse electric field E⃗=Ezk^\vec{E} = E_z \hat{k} until equilibrium is reached: eEz=evdB  ⟹  Ez=vdBe E_z = e v_d B \implies E_z = v_d B

The potential difference VV developed between the two opposite faces (points KK and MM, separated by width ww along the zz-axis) is: V=Ezw=vdBwV = E_z w = v_d B w

Substituting the value of vdv_d: V=(Inewd)Bw=IBnedV = \left(\frac{I}{n e w d}\right) B w = \frac{I B}{n e d}

From this result, we see that for fixed current II and thickness dd: V∝BnV \propto \frac{B}{n}

The ratio of the potential differences V1V_1 and V2V_2 for strips 1 and 2 is: V2V1=(B2B1)(n1n2)\frac{V_2}{V_1} = \left(\frac{B_2}{B_1}\right) \left(\frac{n_1}{n_2}\right)

3. Evaluation of the Options:

  • Option A: If B1=B2B_1 = B_2 and n1=2n2n_1 = 2n_2: V2V1=(1)×(2n2n2)=2  ⟹  V2=2V1\frac{V_2}{V_1} = (1) \times \left(\frac{2n_2}{n_2}\right) = 2 \implies V_2 = 2V_1 (Correct)

  • Option B: If B1=B2B_1 = B_2 and n1=2n2n_1 = 2n_2, then V2=2V1≠V1V_2 = 2V_1 \neq V_1. (Incorrect)

  • Option C: If B1=2B2B_1 = 2B_2 and n1=n2n_1 = n_2: V2V1=(B22B2)×(1)=0.5  ⟹  V2=0.5V1\frac{V_2}{V_1} = \left(\frac{B_2}{2B_2}\right) \times (1) = 0.5 \implies V_2 = 0.5V_1 (Correct)

  • Option D: If B1=2B2B_1 = 2B_2 and n1=n2n_1 = n_2, then V2=0.5V1≠V1V_2 = 0.5V_1 \neq V_1. (Incorrect)

Conclusion:

The correct options are A and C.

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