Question

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities. Determine the radius of their circular path.

Answer 1

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

$r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$

m is mass of Singly charged uranium-238 ion, $m=3.95\times 10^{-25}\ kg$

q is charge

So,

$r=\dfrac{1}{1.9}\times \sqrt{\dfrac{2\times 3.95\times 10^{-25}\times 2.2\times 10^3}{1.6\times 10^{-19}}}\\\\r=0.0548\ m$

So, the radius of their circular path is equal to 0.0548 m.