Question

A deuteron, with the same charge but twice the mass of a proton, moves with a speed of 6.00 × 105 m/s perpendicular to a uniform magnetic field of 0.0525 T. Which of the paths described below would it follow? (qp = 1.60 × 10−19 C and md = 3.34 × 10−27 kg)

Answer 1

Explanation:

The given data is as follows.

     v = $6.00 \times 10^{5} m/s$

     B = 0.0525 T,    q = $1.60 \times 10^{-19}$

     m = $3.34 \times 10^{-27} kg$

It is known that relation between mass and magnetic field is as follows.

          $\frac{mv^{2}}{r} = Bvq$

or,       r = $\frac{mv^{2}}{Bvq}$

So, putting the given values into the above formula and we will calculate the radius as follows.

             r = $\frac{mv^{2}}{Bvq}$

               = $\frac{3.34 \times 10^{-27} kg \times 6.00 \times 10^{5} m/s}{0.0525 T \times 1.60 \times 10^{-19}}$

               = $\frac{20.04 \times 10^{-22}}{0.084 \times 10^{-19}}$

               = 0.238 m

Thus, we can conclude that radius of the circular path is 0.238 m.

Answer 2

Answer:

The radius is $238.57\times10^{-3}\ m$

Explanation:

Given that,

Speed $v=6.00\times10^{5}\ m/s$

Magnetic field = 0.0525 T

We need to calculate the radius

Using relation of centripetal force and magnetic force

$F=qvB$

$\dfrac{mv^2}{r}=qvB$

$r=\dfrac{mv^2}{qvB}$

$r=\dfrac{mv}{qB}$

Put the value into the formula

$r=\dfrac{3.34\times10^{-27}\times6.00\times10^{5}}{1.60\times10^{-19}\times0.0525}$

$r=0.238\ m$

$r=238.57\times10^{-3}\ m$

Hence, The radius is $238.57\times10^{-3}\ m$