JEE Advanced PYQ Physics – Electric Current, Drift of Electrons & Ohm’s Law

Q 1 :

An infinite line charge of uniform electric charge density $λ$ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time $t = 0$ , the space inside the cylinder is filled with a material of permittivity $ε$ and electrical conductivity $σ$ . The electrical conduction in the material follows Ohm's law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density $j (t)$ at any point in the material? [2016]

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[IMAGE 781]
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4

(3)

Electric field at a distance $r$ from a line charge

$E = \frac{λ}{2 π ε r} = \frac{d V}{d r}$

where $λ$ is the linear charge density of the wire.

$d V = - \frac{λ}{2 π ε r} d r$

Current through the elemental shell,

$I = \frac{| d V |}{d R}$ $= \frac{\frac{λ}{2 π ε r} d r}{\frac{1}{σ} \times \frac{d r}{2 π r l}} = \frac{λ σ l}{ε}$ $(∵ R = ρ \frac{l}{A} ∴ d R = ρ \frac{d r}{2 π r l} = \frac{1}{σ} \frac{d r}{2 π r l})$

This current is radially outward.

$∴$ $\frac{d}{d t} (λ l) = - \frac{λ σ l}{ε}$ $\Rightarrow \frac{d λ}{d t} = - \frac{σ}{ε} λ$

Integrating,

$\int_{λ_{0}}^{λ} \frac{d λ}{λ} = - \frac{σ}{ε} \int_{0}^{t} d t$

$\ln (\frac{λ}{λ_{0}}) = - \frac{σ}{ε} t$

$∴$ $λ = λ_{0} e^{- (σ / ε) t}$

Current density,

$J = \frac{I}{2 π r l}$ $= \frac{λ σ l}{2 π r l ε}$ $= \frac{λ σ}{2 π ε r}$

Substituting $λ = λ_{0} e^{- (σ / ε) t}$ ,

$J = (\frac{λ_{0} σ}{2 π ε r}) e^{- (σ / ε) t}$

$∴$ $J = J_{0} e^{- (σ / ε) t}$

Q 2 :

Consider a thin square sheet of side L and thickness $t$ , made of a material of resistivity $ρ$ . The resistance between two opposite faces, shown by the shaded areas in the figure, is [2010]

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directly proportional to $L$
directly proportional to $t$
independent of $L$
independent of $t$

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(3)

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We know that resistance, $R = ρ \frac{l}{a}$

Here, $l = L$ and $a = L \times t$

Therefore, $R = \frac{ρ L}{L \times t} = \frac{ρ}{t}$

Hence, R is independent of L and inversely proportional to $t .$

Q 3 :

To verify Ohm's law, a student is provided with a test resistor $R_{T}$ , a high resistance $R_{1}$ , a small resistance $R_{2}$ , two identical galvanometers $G_{1}$ and $G_{2}$ , and a variable voltage source $V$ . The correct circuit to carry out the experiment is [2010]

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2

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(3)

A voltmeter is made by connecting a high resistance $R_{1}$ in series with the galvanometer $G_{1}$ .

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An ammeter is made by connecting a low resistance $R_{2}$ in parallel with the galvanometer $G_{2}$ .

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The voltmeter is connected in parallel with the test resistor $R_{T}$ in the circuit.

The ammeter is connected in series with the test resistor $R_{T}$ in the circuit.

A variable voltage source $V$ is connected in series with the test resistor $R_{T}$ in the circuit.

Q 4 :

Shown in the figure is a semicircular metallic strip that has thickness $t$ and resistivity $ρ$ . Its inner radius is $R_{1}$ and outer radius is $R_{2}$ . If a voltage $V_{0}$ is applied between its two ends, a current $I$ flows in it. In addition, it is observed that a transverse voltage $Δ V$ develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignoring any role of the magnetic field due to the current). Then (the figure is schematic and not drawn to scale) [2020]

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$I = \frac{V_{0} t}{π ρ} \ln (\frac{R_{2}}{R_{1}})$
The outer surface is at a higher voltage than the inner surface.
The outer surface is at a lower voltage than the inner surface.
$Δ V \propto I^{2}$

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4

Select one or more options

(1, 3, 4)

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Resistance of elementary strips

$\int \frac{1}{d R} = \int_{R_{1}}^{R_{2}} \frac{t d x}{ρ π x}$ $\Rightarrow \frac{1}{R} = \frac{t}{π ρ} \ln (\frac{R_{2}}{R_{1}})$

Therefore, $R = \frac{π ρ}{t \ln (\frac{R_{2}}{R_{1}})}$ $\Rightarrow I = \frac{V_{0}}{R} = \frac{V_{0} t \ln (\frac{R_{2}}{R_{1}})}{π ρ}$

Hence, option (1) is correct.

For the circular motion of electrons, $Δ V$ develops.

The inner surface is at a higher potential, so that the electric field develops radially outward.

Therefore, option (3) is correct.

$\frac{m V_{d}^{2}}{r} = q \vec{E}$ $\Rightarrow \vec{E} = \frac{m V_{d}^{2}}{q r}$

Using

$V_{d} = \frac{I}{n e A},$

we get

$E = \frac{m}{q r} {(\frac{I}{n e A})}^{2} = \frac{m I^{2}}{n^{2} e^{2} A^{2} q r}$

Since

$Δ V = \int \vec{E} \cdot d \vec{r}$ or $Δ V \propto V_{d}^{2}$

$∴$ $Δ V \propto I^{2}$

Therefore, option (4) is correct.

Topic Question Set

Q 2 :

Consider a thin square sheet of side L and thickness $t$ , made of a material of resistivity $ρ$ . The resistance between two opposite faces, shown by the shaded areas in the figure, is [2010]

[IMAGE 784]

Q 3 :

To verify Ohm's law, a student is provided with a test resistor $R_{T}$ , a high resistance $R_{1}$ , a small resistance $R_{2}$ , two identical galvanometers $G_{1}$ and $G_{2}$ , and a variable voltage source $V$ . The correct circuit to carry out the experiment is [2010]

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Topic Question Set

Q 2 : Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure, is [2010] [IMAGE 784]

Q 3 : To verify Ohm's law, a student is provided with a test resistor RT, a high resistance R1, a small resistance R2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct circuit to carry out the experiment is [2010]

Want Us to Email you About Special Offers & Updates?

Q 2 :

Consider a thin square sheet of side L and thickness $t$ , made of a material of resistivity $ρ$ . The resistance between two opposite faces, shown by the shaded areas in the figure, is [2010]

[IMAGE 784]

Q 3 :

To verify Ohm's law, a student is provided with a test resistor $R_{T}$ , a high resistance $R_{1}$ , a small resistance $R_{2}$ , two identical galvanometers $G_{1}$ and $G_{2}$ , and a variable voltage source $V$ . The correct circuit to carry out the experiment is [2010]