Q158 / 180 · COMEDK 2023 (Evening Shift) · Physics
The radius of the current carrying circular coil is doubled keeping the current passing through it the same. Then the ratio of the magnetic field produced at the centre of the coil before the doubling of the radius to the magnetic field after doubling of the radius.
- Correct answerA2 : 1
- B1 : 2
- C2 : 3
- D3 : 2
Solution
The magnetic field $B$ at the centre of a circular coil of radius $R$ carrying current $I$ is given by the formula $B = \dfrac{\mu_{0} I}{2R}$.
Let the initial radius be $R_{1} = R$ and the initial magnetic field be $B_{1} = \dfrac{\mu_{0} I}{2R}$.
After doubling the radius, the new radius is $R_{2} = 2R$. The current $I$ remains the same. The new magnetic field is $B_{2} = \dfrac{\mu_{0} I}{2R_{2}} = \dfrac{\mu_{0} I}{2(2R)} = \dfrac{\mu_{0} I}{4R}$.
The ratio of the magnetic field before doubling to the magnetic field after doubling is $\dfrac{B_{1}}{B_{2}} = \dfrac{\dfrac{\mu_{0} I}{2R}}{\dfrac{\mu_{0} I}{4R}} = \dfrac{4R}{2R} = \dfrac{2}{1}$.
Therefore, the ratio is $2 : 1$.
Answer: 2 : 1
Let the initial radius be $R_{1} = R$ and the initial magnetic field be $B_{1} = \dfrac{\mu_{0} I}{2R}$.
After doubling the radius, the new radius is $R_{2} = 2R$. The current $I$ remains the same. The new magnetic field is $B_{2} = \dfrac{\mu_{0} I}{2R_{2}} = \dfrac{\mu_{0} I}{2(2R)} = \dfrac{\mu_{0} I}{4R}$.
The ratio of the magnetic field before doubling to the magnetic field after doubling is $\dfrac{B_{1}}{B_{2}} = \dfrac{\dfrac{\mu_{0} I}{2R}}{\dfrac{\mu_{0} I}{4R}} = \dfrac{4R}{2R} = \dfrac{2}{1}$.
Therefore, the ratio is $2 : 1$.
Answer: 2 : 1
Hence, the correct answer is option: 'A'.