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Magnetism question

2015 · Shift 2 · Q60
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Magnetism question

2015 · Shift 2 · Q60

JEE AdvancedPhysicsMagnetismMultiple correct+4 / −2
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are l, w and d, respectively. A uniform magnetic field B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. JEE Advanced 2015 Paper 2 Offline Physics - Magnetism Question 37 English ComprehensionConsider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct options is/are :
  1. A
    If B1 = B2 and n1 = 2n2, then V2 = 2V1
  2. B
    If B1 = B2 and n1 = 2n2, then V2 = V1
  3. C
    If B1 = 2B2 and n1 = n2, then V2 = 0.5V1
  4. D
    If B1 = 2B2 and n1 = n2, then V2 = V1
View written solutionFree

Correct answer: A, C

Step 1: Identify the phenomenon

This is the Hall effect.

A current III flows along +x+x+x direction, and magnetic field B⃗\vec BB is along +y+y+y direction. Since the charge carriers are electrons, they experience magnetic deflection along the zzz-direction, causing charge separation and hence a potential difference across the thickness.

Step 2: Magnetic force balanced by electric force

At equilibrium,

∣q∣E=∣q∣vdB|q|E = |q|v_d B∣q∣E=∣q∣vd​B

so,

E=vdBE = v_d BE=vd​B

If the potential difference across the thickness ddd is VVV, then

E=VdE = \frac{V}{d}E=dV​

Hence,

Vd=vdB\frac{V}{d} = v_d BdV​=vd​B V=vdBdV = v_d B dV=vd​Bd

Step 3: Express drift velocity in terms of current

Current in the strip is

I=neAvdI = n e A v_dI=neAvd​

where cross-sectional area perpendicular to current is

A=wdA = wdA=wd

Therefore,

vd=Inewdv_d = \frac{I}{n e w d}vd​=newdI​

Substitute into the Hall voltage expression:

V=(Inewd)BdV = \left(\frac{I}{n e w d}\right)BdV=(newdI​)Bd V=IBnewV = \frac{IB}{n e w}V=newIB​

So Hall voltage is

V∝BnV \propto \frac{B}{n}V∝nB​

for fixed I,e,wI, e, wI,e,w.

Step 4: Compare the two strips

For strip 1:

V1=IB1n1ewV_1 = \frac{I B_1}{n_1 e w}V1​=n1​ewIB1​​

For strip 2:

V2=IB2n2ewV_2 = \frac{I B_2}{n_2 e w}V2​=n2​ewIB2​​

Thus,

V2V1=B2/n2B1/n1=B2n1B1n2\frac{V_2}{V_1} = \frac{B_2/n_2}{B_1/n_1} = \frac{B_2 n_1}{B_1 n_2}V1​V2​​=B1​/n1​B2​/n2​​=B1​n2​B2​n1​​

Step 5: Check each option

Option A

Given:

B1=B2,n1=2n2B_1 = B_2, \qquad n_1 = 2n_2B1​=B2​,n1​=2n2​

Then,

V2V1=B2n1B1n2=n1n2=2\frac{V_2}{V_1} = \frac{B_2 n_1}{B_1 n_2} = \frac{n_1}{n_2} = 2V1​V2​​=B1​n2​B2​n1​​=n2​n1​​=2

So,

V2=2V1V_2 = 2V_1V2​=2V1​

Option A is correct.

Option B

Same condition as above gives

V2=2V1V_2 = 2V_1V2​=2V1​

not V2=V1V_2 = V_1V2​=V1​.

Option B is incorrect.

Option C

Given:

B1=2B2,n1=n2B_1 = 2B_2, \qquad n_1 = n_2B1​=2B2​,n1​=n2​

Then,

V2V1=B2B1=12\frac{V_2}{V_1} = \frac{B_2}{B_1} = \frac{1}{2}V1​V2​​=B1​B2​​=21​

So,

V2=0.5V1V_2 = 0.5V_1V2​=0.5V1​

Option C is correct.

Option D

From above,

V2=0.5V1V_2 = 0.5V_1V2​=0.5V1​

not V2=V1V_2 = V_1V2​=V1​.

Option D is incorrect.

Final Answer

Correct options are:

A, C\boxed{A,\ C}A, C​

Comparison with stored correct answer

Stored correct answer: A, C

My derived answer matches the stored answer.

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