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A deuteron (the nucleus of an isotope of hydrogen) has a mass of $3.34 \times 10^{- 27}$ kg and a charge of ve. The deuteron travels in a circular path with a radius of $6.96$ mm in a magnetic field with magnitude $2.50 T$ . Through what potential difference would the deuteron have to be accelerated to acquire this speed?

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$V = \frac{r^{2} B^{2} q}{2 m}$

$V = \frac{r^{2} B^{2} q}{m}$

$V = \frac{r B^{2} q}{2 m}$

$V = \frac{r^{2} B q}{2 m}$

answer is A.

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Detailed Solution

$r = \frac{\sqrt{2 q V m}}{B q} ∴ V = \frac{r^{2} B^{2} q^{2}}{2 q m} = \frac{r^{2} B^{2} q}{2 m}$

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