Magnetics 2 Question 11
11. A long straight wire along the $z$-axis carries a current $I$ in the negative $z$-direction. The magnetic vector field $\mathbf{B}$ at a point having coordinate $(x, y)$ on the $z=0$ plane is
(2002, 2M)
(a) $\frac{\mu _0 I(y \hat{\mathbf{i}}-x \hat{\mathbf{j}})}{2 \pi\left(x^{2}+y^{2}\right)}$
(b) $\frac{\mu _0 I(x \hat{\mathbf{i}}+y \hat{\mathbf{j}})}{2 \pi\left(x^{2}+y^{2}\right)}$
(c) $\frac{\mu _0 I(x \hat{\mathbf{j}}-y \hat{\mathbf{i}})}{2 \pi\left(x^{2}+y^{2}\right)}$
(d) $\frac{\mu _0 I(x \hat{\mathbf{i}}-y \hat{\mathbf{j}})}{2 \pi\left(x^{2}+y^{2}\right)}$
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Answer:
Correct Answer: 11. (a)
Solution:
- Magnetic field at $P$ is $\mathbf{B}$, perpendicular to $O P$ in the direction shown in figure.
So, $\quad \mathbf{B}=B \sin \theta \hat{\mathbf{i}}-B \cos \theta \hat{\mathbf{j}}$
Here, $\quad B=\frac{\mu _0 I}{2 \pi r}$
$$ \begin{aligned} \sin \theta & =\frac{y}{r} \quad \text { and } \quad \cos \theta=\frac{x}{r} \\ \quad \mathbf{B} & =\frac{\mu _0 I}{2 \pi} \cdot \frac{1}{r^{2}}(y \hat{\mathbf{i}}-x \hat{\mathbf{i}}) \\ & =\frac{\mu _0 I(y \hat{\mathbf{i}}-x \hat{\mathbf{j}})}{2 \pi\left(x^{2}+y^{2}\right)} \quad\left(\text { as } r^{2}=x^{2}+y^{2}\right) \end{aligned} $$
