A long current carrying wire produces a magnetic field of 1 $\mathrm{T}$ at a distance of $r$. The magnetic…

A long current carrying wire produces a magnetic field of 1 $\mathrm{T}$ at a distance of $r$. The magnetic field (a) $\frac{\mathrm{r}}{2}$ (b) $2 \mathrm{r}$ and (c) $3 \mathrm{r}$ is
  1. (a) $2 \mathrm{~T},(\mathrm{~b})=\frac{1}{2} \mathrm{~T},(\mathrm{c})=\frac{1}{3} \mathrm{~T}$
  2. (a) $3 \mathrm{~T},(\mathrm{~b})=\frac{1}{3} \mathrm{~T},(\mathrm{c})=\frac{1}{6} \mathrm{~T}$
  3. (a) $\frac{3}{2} \mathrm{~T},(\mathrm{~b})=\frac{1}{4} \mathrm{~T},(\mathrm{c})=\frac{1}{8} \mathrm{~T}$
  4. (a) $\frac{5}{2} \mathrm{~T},\left(\right.$ b) $=\frac{1}{2} \mathrm{~T},(\mathrm{c})=\frac{1}{3} \mathrm{~T}$

Solution

For a long straight wire $ B=\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}} $ $\mathrm{B} \propto \frac{1}{\mathrm{r}} \Rightarrow \frac{\mathrm{B}_{\mathrm{r} / 2}}{\mathrm{~B}_{\mathrm{r}}}=\frac{\mathrm{r}}{\frac{\mathrm{r}}{2}} \Rightarrow \frac{\mathrm{B}_{\mathrm{r} / 2}}{\mathrm{~B}_{\mathrm{r}}}=2 \Rightarrow \mathrm{B}_{\mathrm{r} / 2}=2 \times 1=2 \mathrm{~T}$ Similarly, $\mathrm{B}_{2 \mathrm{r}}=\frac{1}{2} \mathrm{~T}$ and $\mathrm{B}_{3 \mathrm{r}}=\frac{1}{3} \mathrm{~T}$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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