If an electron is moving in a magnetic field of $5.4\times {10}^{-4} T$ on a circular path of radius $32 cm$ having a frequency of $2.5 MHz,$then its speed will be

Here, $$\begin{gathered} B=5.4\times {10}^{-4} T; \\ r=32 cm=32\times {10}^{-2} m; \nu =2.5 MHz=2.5\times {10}^6 Hz \end{gathered}$$

The speed of electron on circular path 

$$\begin{gathered} \nu =r\times 2\mathrm{\pi \nu }=32\times {10}^{-2}\times 2\times 3.14\times 2.5\times {10}^6 \\ =502.4\times {10}^4=5.024\times {10}^6 {\mathrm{ms}}^{-1} \end{gathered}$$