Question

If the radius of a circular coil is doubled and the number of turns is halved, then the magnetic field at the centre of the coil for the same current is

• A. becomes doubled
• B. becomes one fourth
• C. becomes quadrupled
• D. remains unchanged

Hint:

For magnetic field calculations, always pay attention to the direct and inverse proportionality of the formula. In this case, $B \propto N$ (directly proportional to turns) and $B \propto 1/r$ (inversely proportional to radius). Apply the changes to these proportionalities to quickly assess the overall effect.

Answer 1

The Correct Option is B

Step 1: Recall the Formula for Magnetic Field at the Centre of a Circular Coil
The magnetic field ($B$) at the centre of a circular coil with $N$ turns, radius $r$, and carrying a current $I$ is given by the formula: \[ B = \frac{\mu_0 N I}{2r} \] Where $\mu_0$ is the permeability of free space (a constant). Step 2: Identify the Initial Conditions
Let the initial magnetic field be $B_1$.
Initial number of turns = $N_1 = N$
Initial radius = $r_1 = r$
Initial current = $I_1 = I$
So, the initial magnetic field is: \[ B_1 = \frac{\mu_0 N I}{2r} \quad \cdots (1) \] Step 3: Identify the New Conditions
The problem states that the radius of the circular coil is doubled and the number of turns is halved, while the current remains the same.
New number of turns = $N_2 = \frac{N}{2}$
New radius = $r_2 = 2r$
New current = $I_2 = I$
Step 4: Calculate the New Magnetic Field
Let the new magnetic field be $B_2$. Substitute the new conditions into the formula for $B$: \[ B_2 = \frac{\mu_0 N_2 I_2}{2r_2} \] \[ B_2 = \frac{\mu_0 \left(\frac{N}{2}\right) I}{2(2r)} \] \[ B_2 = \frac{\mu_0 N I}{4 \times 2r} \] \[ B_2 = \frac{\mu_0 N I}{8r} \quad \cdots (2) \] Step 5: Compare the New Magnetic Field with the Initial Magnetic Field
Divide equation (2) by equation (1): \[ \frac{B_2}{B_1} = \frac{\frac{\mu_0 N I}{8r}}{\frac{\mu_0 N I}{2r}} \] \[ \frac{B_2}{B_1} = \frac{1/8}{1/2} \] \[ \frac{B_2}{B_1} = \frac{1}{8} \times 2 \] \[ \frac{B_2}{B_1} = \frac{1}{4} \] Therefore, $B_2 = \frac{1}{4} B_1$. The magnetic field becomes one fourth of its initial value. Step 6: Analyze Options
\begin{itemize} \item Option (1): becomes doubled. Incorrect. \item Option (2): becomes one fourth. Correct, as it matches our calculated result. Option (3): becomes quadrupled. Incorrect. \item Option (4): remains unchanged. Incorrect. \end{itemize}