Q136 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
The ratio of magnetic field at the centre of the current carrying circular loop and its magnetic moment is '$x$'. When both the current and radius are tripled, then the ratio will be
- Correct answerA$\dfrac{x}{27}$
- B$\dfrac{x}{9}$
- C$\dfrac{x}{3}$
- D$3x$
Solution
Magnetic field at the centre of a circular loop is $B = \dfrac{\mu_0 I}{2R}$
Magnetic moment of the loop is $M = I \pi R^2$
The ratio is $x = \dfrac{B}{M} = \dfrac{\mu_0 I}{2R \cdot I \pi R^2} = \dfrac{\mu_0}{2 \pi R^3}$
When current and radius are tripled, $I' = 3I$ and $R' = 3R$
New magnetic field $B' = \dfrac{\mu_0 (3I)}{2(3R)} = \dfrac{\mu_0 I}{2R} = B$
New magnetic moment $M' = (3I) \pi (3R)^2 = 27 I \pi R^2 = 27 M$
New ratio $x' = \dfrac{B'}{M'} = \dfrac{B}{27M} = \dfrac{x}{27}$
Answer: $\dfrac{x}{27}$
Magnetic moment of the loop is $M = I \pi R^2$
The ratio is $x = \dfrac{B}{M} = \dfrac{\mu_0 I}{2R \cdot I \pi R^2} = \dfrac{\mu_0}{2 \pi R^3}$
When current and radius are tripled, $I' = 3I$ and $R' = 3R$
New magnetic field $B' = \dfrac{\mu_0 (3I)}{2(3R)} = \dfrac{\mu_0 I}{2R} = B$
New magnetic moment $M' = (3I) \pi (3R)^2 = 27 I \pi R^2 = 27 M$
New ratio $x' = \dfrac{B'}{M'} = \dfrac{B}{27M} = \dfrac{x}{27}$
Answer: $\dfrac{x}{27}$
Hence, the correct answer is option: 'A'.