Current I enters at point A and leaves at point D of a conducting block. To find the voltage V between points B and C, we assume current spreads over - Physics MCQ AIEEE 2026 | ExamTest.live

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Current I enters at point A and leaves at point D of a conducting block. To find the voltage Δ V between points B and C, we assume current spreads over a hemispherical surface. The electric field at distance r is E = ρ j . The potential difference V_{B} - V_{C} is:

Solution & Step-by-step Explanation

Current density j = \frac{I}{Area} = \frac{I}{2 π r ^{2}} (Hemispherical surface).Electric field E = ρ j = \frac{ρ I}{2 π r ^{2}} .Potential V = - \int E d r = - \int \frac{ρ I}{2 π r ^{2}} d r = \frac{ρ I}{2 π r} .The potential difference due to one source: V_{B} - V_{C} = \frac{ρ I}{2 πa} - \frac{ρ I}{2 π ( a + b )} = \frac{ρ I}{2 π} (\frac{1}{a} - \frac{1}{a + b}) .

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Current I enters at point A and leaves at point D of a conducting block. To find the voltage Δ V between points B and C, we assume current spreads over a hemispherical surface. The electric field at distance r is E = ρ j . The potential difference V_{B} - V_{C} is:

\frac{ρ I}{π} (\frac{1}{a} - \frac{1}{a + b})

ρ I (\frac{1}{a} - \frac{1}{a + b})

\frac{ρ I}{2 π} (\frac{1}{a} - \frac{1}{a + b})

\frac{ρ I}{4 π} (\frac{1}{a} - \frac{1}{a + b})

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