Question

A 6.00 T magnetic field is applied perpendicular to the path of charged particles in a bubble chamber. What is the radius of curvature (in m) of the path of a 14.4 MeV proton in this field

Answer 1

The radius of curvature of the path of a 14.4 MeV proton in a 6.00 T magnetic field is 0.091 m.

The radius of curvature can be calculated with Lorentz force:

$F = q(E + v\times B)$

Since there is no electric field (E = 0) and the Lorentz force is equal to the centripetal force, we have:

$ma_{c} = q(v\times B)$

Replacing $a_{c}$ for $frac{v^{2}}{r}$ in the above equation, we get:

$\frac{mv^{2}}{r} = q(v\times B)$

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

q: is the charge of the proton = 1.602x10⁻¹⁹ C

r: is the radius of curvature =?

v: is the tangential velocity

B: is the magnetic field = 6.00 T

The magnetic field is perpendicular to the charged particles, so:

$\frac{mv^{2}}{r} = q[vBsin(90)]$

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

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

The tangential velocity can be calculated from the energy:

$E = \frac{1}{2}mv^{2}$

$v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2*14.4 \cdot 10^{6} eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}}{1.67 \cdot 10^{-27} kg}} = 5.26 \cdot 10^{7} m/s$

Finally, the radius is:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*5.26 \cdot 10^{7} m/s}{1.602 \cdot 10^{-19} C*6.00 T} = 0.091 m$  

Therefore, the radius of curvature of the path is 0.091 m.

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