JEE Advance - Physics (2024 - Paper 1 Online - No. 2)
An infinitely long wire, located on the $z$-axis, carries a current $I$ along the $+z$-direction and produces the magnetic field $\vec{B}$. The magnitude of the line integral $\int \vec{B} \cdot \overrightarrow{d l}$ along a straight line from the point $(-\sqrt{3} a, a, 0)$ to $(a, a, 0)$ is given by
[ $\mu_0$ is the magnetic permeability of free space.]
Explanation
$$\begin{aligned} & \Rightarrow|\overrightarrow{\mathrm{d} \ell}|=\mathrm{rd} \theta \\ & \Rightarrow|\overrightarrow{\mathrm{B}}|=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=\int|\overline{\mathrm{B}}||\overrightarrow{\mathrm{d} \ell}| \cos 0^{\circ} \\ & =\int\left(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}\right) \times(\mathrm{rd} \theta) \\ & =\int_\limits{\theta_1}^{\theta_2} \frac{\mu_0 \mathrm{I}}{2 \pi} \mathrm{d} \theta=\frac{\mu_0 \mathrm{I}}{2 \pi}\left[\theta_2-\left(-\theta_1\right)\right] \end{aligned}$$
$$\text { [ } \theta_1 \text { is anticlockwise hence taken negative] }$$
$$\begin{aligned} & \Rightarrow \tan \theta_1=\frac{a \sqrt{3}}{a}=\sqrt{3} \\ & \theta_1=\frac{\pi}{3} \\ & \Rightarrow \tan \theta_2=\frac{a}{a}=1 \\ & \theta_2=\frac{\pi}{4} \\ & \Rightarrow \int B \cdot d \ell=\frac{\mu_0 I}{2 \pi}\left[\frac{\pi}{3}+\frac{\pi}{4}\right] \\ & =\frac{7 \mu_0 I}{24} \\ & \Rightarrow \text { Ans. Option (A) } \end{aligned}$$
Comments (0)
