Magnetic Field of a Cylindrical Conductor – Rankers Physics

Magnetic Field Due to Straight Current Carrying Wire: Practice Problem & Solution

Assertion (A): The magnetic field induction due to an infinite long current carrying solid cylindrical conductor of radius \(R\), at a distance \(R/2\) and \(2R\) from its axis is same. Reason (R): An infinite long current carrying solid cylindrical conductor is a source of uniform magnetic field.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Magnetic Field Due to Straight Current Carrying Wire. Understanding the underlying formula is key to arriving at the correct answer below:

Assertion (A) is true: Using Ampere's Law, \(B(R/2) = \frac{\mu_0 I (R/2)}{2\pi R^2} = \frac{\mu_0 I}{4\pi R}\) and \(B(2R) = \frac{\mu_0 I}{2\pi (2R)} = \frac{\mu_0 I}{4\pi R}\).


Reason (R) is false: The magnetic field is not uniform; it varies linearly inside (\(B \propto r\)) and inversely outside (\(B \propto 1/r\)). Thus, A is true and R is false.

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