Magnetic Effects of Current — JEE Advanced · 2015 · 24 May 2015 · Paper 2 Question with Solution

PARAGRAPH 1

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 to 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.

Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are

w_{1}

and

w_{2}

and thicknesses are

d_{1}

and

d_{2}

, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure).

V_{1}

and

V_{2}

are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is(are)

(A)If $w_{1} = w_{2}$ and $d_{1} = 2 d_{2}$ , then $V_{2} = 2 V_{1}$ ✓ Correct
(B)If $w_{1} = w_{2}$ and $d_{1} = 2 d_{2}$ , then $V_{2} = V_{1}$
(C)If $w_{1} = 2 w_{2}$ and $d_{1} = d_{2}$ , then $V_{2} = 2 V_{1}$
(D)If $w_{1} = 2 w_{2}$ and $d_{1} = d_{2}$ , then $V_{2} = V_{1}$ ✓ Correct

Answer: (A), (D)

Solution

This is the Hall effect. At equilibrium, the magnetic force on electrons equals the electric force:

e v B = e E = e \frac{V}{w}

where

v

is the drift velocity and

V

is the Hall voltage across width

w

.

So

V = w v B

.

The current

I = n e A v = n e (w d) v

, where

n

is the charge carrier density and

A = w d

is the cross-sectional area.

From this:

w v = \frac{I}{n e d}

Therefore:

V = \frac{I B}{n e d}

The Hall voltage is inversely proportional to thickness

d

and independent of width

w

.

(A) If

w_{1} = w_{2}

and

d_{1} = 2 d_{2}

\frac{V _{2}}{V _{1}} = \frac{d _{1}}{d _{2}} = 2

, so

V_{2} = 2 V_{1}

✓

(B) Same conditions give

V_{2} = 2 V_{1}

, not

V_{2} = V_{1}

✗

(C) If

w_{1} = 2 w_{2}

and

d_{1} = d_{2}

: Since

V

is independent of

w

V_{1} = V_{2}

, not

V_{2} = 2 V_{1}

✗

(D) Same conditions give

V_{2} = V_{1}

✓

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Magnetic Effects of Current

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