Q29. A long wire carrying current I is bent in the form of a circle as shown in the figure.
I
O
The

Q: What is the correct answer for CUET UG Physics Q29?

The correct answer is D. The diagram shows a full circular loop of radius 'r' carrying current 'I' in a counter-clockwise direction. According to the right-hand thumb rule, the magnetic field at the center 'O' due to a full circular loop carrying current 'I' is given by B_loop = μ₀I / (2r), and its direction is perpendicularly outward from the plane of the loop. However, the provided options contain terms like (μ₀I/4πr)

Q: Q29. A long wire carrying current I is bent in the form of a circle as shown in the figure. I O The B at point O is: (a) (μ₀I/4πr)(1 – π) perpendicularly outward (b) (μ₀I/2πr) perpendicularly…

The correct answer is D. The diagram shows a full circular loop of radius 'r' carrying current 'I' in a counter-clockwise direction. According to the right-hand thumb rule, the magnetic field at the center 'O' due to a full circular loop carrying current 'I' is given by B_loop = μ₀I / (2r), and its direction is perpendicularly outward from the plane of the loop. However, the provided options contain terms like (μ₀I/4πr)

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

A. (μ₀I/4πr)(1 – π) perpendicularly outward · B. (μ₀I/2πr) perpendicularly inward · C. (μ₀I/4πr)(1+π) perpendicularly inward · D. (μ₀I/4πr)(1 + π) perpendicularly outward

Question

Q29. A long wire carrying current I is bent in the form of a circle as shown in the figure.
I
O
The B at point O is:
(a) (μ₀I/4πr)(1 – π) perpendicularly outward
(b) (μ₀I/2πr) perpendicularly inward
(c) (μ₀I/4πr)(1+π) perpendicularly inward
(d) (μ₀I/4πr)(1 + π) perpendicularly outward

FirstInTest · Accepted Answer

The diagram shows a full circular loop of radius 'r' carrying current 'I' in a counter-clockwise direction. According to the right-hand thumb rule, the magnetic field at the center 'O' due to a full circular loop carrying current 'I' is given by B_loop = μ₀I / (2r), and its direction is perpendicularly outward from the plane of the loop. 
However, the provided options contain terms like (μ₀I/4πr) and (μ₀I/4r) in various combinations. If we are to assume option (d) is correct, it has the form (μ₀I/4πr) + (μ₀I/4r). The term μ₀I/(4r) corresponds to the magnetic field at the center of a *semicircle* of radius 'r'. The term μ₀I/(4πr) corresponds to the magnetic field at the end of a *semi-infinite straight wire* or a specific finite segment. The diagram, depicting a simple full circular loop with no explicit straight wire segments contributing to the field at the center (a straight wire passing through the center creates zero field at the center), does not directly lead to this formula for a standard interpretation. If the question intended a semicircle (μ₀I/4r) and two semi-infinite straight wires (2 * μ₀I/4πr = μ₀I/2πr), it would be (μ₀I/4r + μ₀I/2πr) = (μ₀I/4πr)*(π+2). 
Given the structure of option (d) as (μ₀I/4πr)(1 + π) = μ₀I/(4πr) + μ₀I/(4r), this suggests a summation of a semi-infinite straight wire's field (μ₀I/(4πr)) and a semicircle's field (μ₀I/(4r)), both contributing outward based on a counter-clockwise current. The actual diagram of a full circle with current I produces B = μ₀I/(2r) outward. There is a discrepancy between the given diagram, standard physics formulas, and the options/solution. Assuming the chosen answer (d) is correct, and the 'outward' direction from the diagram's current sense is also correct, the magnitude implies a non-standard combination of components not clearly depicted. For a full circle, μ₀I/(2r) = (μ₀I/4πr)(2π). Option (d) is (μ₀I/4πr)(1+π). These are clearly different. Given the provided solution indicates 'd', it is chosen, but acknowledging the discrepancy in its derivation from the typical diagram interpretation.

Answer

(μ₀I/4πr)(1 – π) perpendicularly outward

Answer

(μ₀I/2πr) perpendicularly inward

Answer

(μ₀I/4πr)(1+π) perpendicularly inward

Answer

(μ₀I/4πr)(1 + π) perpendicularly outward

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