Q145 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
The current through the 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 strength, then:
- A$B_2 = 4B_1$
- B$B_2 = \dfrac{B_1}{2}$
- C$B_2 = 2B_1$
- Correct answerD$B_2 = \dfrac{B_1}{4}$
Solution
The magnetic field at the center of a circular coil is given by $B = \dfrac{\mu_0 I}{2R}$.
Initial magnetic field is $B_1 = \dfrac{\mu_0 I}{2R}$.
The current is halved, so $I' = \dfrac{I}{2}$.
The radius is doubled, so $R' = 2R$.
Final magnetic field is $B_2 = \dfrac{\mu_0 I'}{2R'} = \dfrac{\mu_0 \left(\dfrac{I}{2}\right)}{2(2R)} = \dfrac{1}{4} \left(\dfrac{\mu_0 I}{2R}\right)$.
Therefore, $B_2 = \dfrac{B_1}{4}$.
Answer: $B_2 = \dfrac{B_1}{4}$
Initial magnetic field is $B_1 = \dfrac{\mu_0 I}{2R}$.
The current is halved, so $I' = \dfrac{I}{2}$.
The radius is doubled, so $R' = 2R$.
Final magnetic field is $B_2 = \dfrac{\mu_0 I'}{2R'} = \dfrac{\mu_0 \left(\dfrac{I}{2}\right)}{2(2R)} = \dfrac{1}{4} \left(\dfrac{\mu_0 I}{2R}\right)$.
Therefore, $B_2 = \dfrac{B_1}{4}$.
Answer: $B_2 = \dfrac{B_1}{4}$
Hence, the correct answer is option: 'D'.