In the $\text{xy - plane}$, the region $y>0$ has a uniform magnetic field $B_1\widehat{k}$ and the region $y<0$ has another uniform magnetic field $B_2\widehat{k}$. A positively charged particle is projected from the origin along the positive $\text{y}-\text{axis}$ with speed $v_0=\pi \mathrm{m}{s}^{-1}$ at $t=0$, as shown in the figure. Neglect gravity in this problem. Let $t=T$ be the time when the particle crosses the $\text{x}-\text{axis}$ from below for the first time. If $B_2=4B_1$, the average speed of the particle, in $\mathrm{m}{\mathrm{s}}^{-1}$, along the $\text{x}-\text{axis}$ in the time interval $T$ is ________.

Average speed along the \(\mathrm{x}\)-axis
\(\left(\mathrm{V}_{\mathrm{x}}\right)=\frac{\int\left|\overrightarrow{\mathrm{V}}_{\mathrm{x}}\right| \mathrm{dt}}{\int \mathrm{dt}}=\frac{\mathrm{d}_1+\mathrm{d}_2}{\mathrm{t}_1+\mathrm{t}_2} \rightarrow(1)\)
We also have,
\(\begin{aligned}
&\mathrm{r}_1=\frac{\mathrm{mv}}{\mathrm{qB}_1}, \mathrm{r}_2=\frac{\mathrm{mv}}{\mathrm{qB} \mathrm{B}_2} \\
&\text { since } \mathrm{B}_1=\frac{\mathrm{B}_2}{4}
\end{aligned}\)
\(\therefore \mathrm{r}_1=4 \mathrm{r}_2 \rightarrow(2)\)
Time in \(B_1 \Rightarrow \frac{\pi B}{q B_1}=t_1\)
Time in \(\mathrm{B}_2 \Rightarrow \frac{\pi \mathrm{B}}{\mathrm{qB}_2}=\mathrm{t}_2\)
Total distance along \(\mathrm{x}\)-axis \(\mathrm{d}_1+\mathrm{d}_2=2 \mathrm{r}_1+2 \mathrm{r}_2=2\left(\mathrm{r}_1+\mathrm{r}_2\right)=2\left(5 \mathrm{r}_2\right)\)
Total time \(\mathrm{T}=\mathrm{t}_1+\mathrm{t}_2=5 \mathrm{t}_2\)
\(\therefore \text { Average speed }=\frac{10 \mathrm{r}_2}{5 \mathrm{t}_2}=2 \frac{\mathrm{mv}}{\mathrm{qB}_2} \times \frac{\mathrm{qB}_2}{\pi \mathrm{m}}=2\)