Question

The current through a circular coil is halved and the radius of the coil is doubled. If $B_1$ and $B_2$ are respectively the initial and final magnetic field strengths at the center of the coil, then:

• A. $B_2 = \frac{B_1}{2}$
• B. $B_2 = 4B_1$
• C. $B_2 = 2B_1$
• D. $B_2 = \frac{B_1}{4}$

Hint:

Since magnetic field strength is directly proportional to current and inversely proportional to radius, halving the current cuts the field in half ($\frac{1}{2}$), and doubling the radius cuts it in half again ($\frac{1}{2}$). Combining these updates gives $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the original strength!

Answer 1

The Correct Option is D

Concept: The magnitude of the magnetic field strength ($B$) at the exact center of a circular current-carrying loop containing $N$ turns is given by the Biot-Savart expression: \[ B = \frac{\mu_0 N I}{2r} \implies B \propto \frac{I}{r} \]

Step 1: Set up the proportionality ratio based on the modified parameters.
Let the initial configuration terms be current $I_1 = I$ and radius $r_1 = r$. The problem states that:
• New current, $I_2 = \frac{I_1}{2} = \frac{I}{2}$
• New radius, $r_2 = 2r_1 = 2r$

Step 2: Evaluate the field strength ratio.
Using the proportionality tracking relation: \[ \frac{B_2}{B_1} = \left(\frac{I_2}{I_1}\right) \times \left(\frac{r_1}{r_2}\right) \] Substituting our parameter values: \[ \frac{B_2}{B_1} = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} \implies B_2 = \frac{B_1}{4} \]