Question

The magnitude of a magnetic field at the centre of a circular coil of radius R, having N turns and carrying a current I can be doubled by changing

• A. \( I \) to \( 2I \) and \( N \) to \( 2N \) keeping \( R \) unchanged
• B. \( N \) to \( \frac{N}{2} \) and keeping \( I \) and \( R \) unchanged
• C. \( N \) to \( 2N \) and \( R \) to \( 2R \) keeping \( I \) unchanged
• D. \( R \) to \( 2R \) and \( I \) to \( 2I \) keeping \( N \) unchanged
• E. \( I \) to \( 2I \) and keeping \( N \) and \( R \) unchanged

Hint:

To double a value in a fraction, you can either double the numerator or halve the denominator. Increasing both by the same factor (like in option C and D) results in no net change.

Answer 1

Answer: (E)

Concept: The magnetic field \(B\) at the center of a circular coil of radius \(R\) with \(N\) turns carrying current \(I\) is given by the formula: \[ B = \frac{\mu_0 N I}{2R} \] where \(\mu_0\) is the permeability of free space. From this formula, we can see that:
• \(B \propto N\) (Number of turns)
• \(B \propto I\) (Current)
• \(B \propto \frac{1}{R}\) (Inversely proportional to radius)

Step 1: Analyze the requirement to double the field.
We want the new field \(B'\) to be \(2B\). This means: \[ \frac{\mu_0 N' I'}{2R'} = 2 \left( \frac{\mu_0 N I}{2R} \right) \]

Step 2: Evaluate Option (E).
If we change \(I\) to \(2I\) while keeping \(N\) and \(R\) constant: \[ B' = \frac{\mu_0 N (2I)}{2R} = 2 \left( \frac{\mu_0 N I}{2R} \right) = 2B \] This correctly doubles the field.

Step 3: Verify why other options are incorrect.

• (A) Doubling \(I\) and \(N\) makes the field \(4B\).
• (C) Doubling \(N\) and \(R\) results in \(B' = \frac{2}{2}B = B\) (No change).
• (D) Doubling \(R\) and \(I\) results in \(B' = \frac{2}{2}B = B\) (No change).