Q135 / 150 · MHT CET 2024 (15 May Shift 1) · Physics

The strength of magnetic field at a perpendicular distance ' $x$ ' near a long straight conductor carrying current ' $I$ ' is ' $B$ '. The magnetic. field at a distance $\frac{\mathrm{X}}{3}$ from straight conductor will be

  • A
    $\frac{B}{3}$
  • Correct answer
    B
    3 B
  • C
    $\frac{\mathrm{B}^2}{9}$
  • D
    $9 \mathrm{~B}^2$

Solution

$\begin{array}{ll} & \mathrm{B} \propto \frac{1}{\mathrm{r}} \\ & \therefore \\ \therefore & \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mathrm{r}_2}{\mathrm{r}_1} \\ \therefore & \frac{\mathrm{~B}}{\mathrm{~B}_2}=\frac{\frac{\mathrm{x}}{3}}{\mathrm{x}} \\ \therefore & \frac{\mathrm{B}}{\mathrm{B}_2}=\frac{1}{3} \\ \therefore & \mathrm{~B}_2=3 \mathrm{~B}\end{array}$

Hence, the correct answer is option: 'B'.

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