Q11. A long straight wire of circular cross section with radius 'a' carries steady current 'I' which

Q: What is the correct answer for CUET UG Physics Q11?

The correct answer is d. To determine the magnetic field (B) for a long straight wire of radius 'a' carrying a uniformly distributed current 'I', we use Ampere's Circuital Law. 1. **Inside the wire (r < a):** The current enclosed by an Amperian loop of radius 'r' is I_enclosed = (πr²/πa²) * I = Ir²/a². Applying Ampere's Law, ∮ B ⋅ dl = μ₀I_enclosed, we get B(2πr) = μ₀(Ir²/a²), which simplifies to B = (μ₀Ir)/(2πa²). This

Q: Q11. A long straight wire of circular cross section with radius 'a' carries steady current 'I' which is uniformly distributed across the cross-section. The magnetic field in the region r …

The correct answer is d. To determine the magnetic field (B) for a long straight wire of radius 'a' carrying a uniformly distributed current 'I', we use Ampere's Circuital Law. 1. **Inside the wire (r < a):** The current enclosed by an Amperian loop of radius 'r' is I_enclosed = (πr²/πa²) * I = Ir²/a². Applying Ampere's Law, ∮ B ⋅ dl = μ₀I_enclosed, we get B(2πr) = μ₀(Ir²/a²), which simplifies to B = (μ₀Ir)/(2πa²). This

Q: What are the options for this CUET UG Physics question?

A. B (graph shows B increasing linearly from 0 at r=0 to a max at r=a, then decreasing as 1/r for r>a) · B. B (graph shows B constant for r a) · C. B (graph shows B decreasing linearly from a max at r=0 to 0 at r=a, then decreasing as 1/r for r>a) · D. B (graph shows B constant for r a)

Question

Q11. A long straight wire of circular cross section with radius 'a' carries steady current 'I' which is uniformly distributed across the cross-section. The magnetic field in the region r < a and r > a is represented by: (a) B (graph shows B increasing linearly from 0 at r=0 to a max at r=a, then decreasing as 1/r for r>a) (b) B (graph shows B constant for ra) (c) B (graph shows B decreasing linearly from a max at r=0 to 0 at r=a, then decreasing as 1/r for r>a) (d) B (graph shows B constant for ra)

FirstInTest · Accepted Answer

To determine the magnetic field (B) for a long straight wire of radius 'a' carrying a uniformly distributed current 'I', we use Ampere's Circuital Law.
1.  **Inside the wire (r < a):** The current enclosed by an Amperian loop of radius 'r' is I_enclosed = (πr²/πa²) * I = Ir²/a². Applying Ampere's Law, ∮ B ⋅ dl = μ₀I_enclosed, we get B(2πr) = μ₀(Ir²/a²), which simplifies to B = (μ₀Ir)/(2πa²). This shows that B is directly proportional to 'r' (B ∝ r) and increases linearly from the center (r=0) to the surface (r=a).
2.  **Outside the wire (r > a):** The current enclosed by an Amperian loop of radius 'r' is the total current 'I'. Applying Ampere's Law, B(2πr) = μ₀I, which gives B = (μ₀I)/(2πr). This shows that B is inversely proportional to 'r' (B ∝ 1/r) and decreases hyperbolically with distance. The combination of these two behaviors results in a graph where B increases linearly inside the wire and decreases as 1/r outside. The visual description in option (a) matches this physical behavior. However, the provided solution marks (d) as correct. If option (d) represents the physically correct graph (linear increase inside, 1/r decrease outside), then the OCR description of (d) in the question text is inaccurate, or the solution's 'Ans (d)' refers to a graph that visually matches option (a)'s description.

Answer

B (graph shows B increasing linearly from 0 at r=0 to a max at r=a, then decreasing as 1/r for r>a)

Answer

B (graph shows B constant for r<a, then decreasing as 1/r for r>a)

Answer

B (graph shows B decreasing linearly from a max at r=0 to 0 at r=a, then decreasing as 1/r for r>a)

Answer

B (graph shows B constant for r<a, then constant for r>a)

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