A deuteron is moving with a speed of 2.0 $\times$ 10⁶ m/s at right angles to a magnetic field.The field is uniform,with magnitude B = 0.40 T.The mass and charge of a deuteron are 3.3 $\times$ 10^-27 kg and 1.6 $\times$ 10^-19 C,respectively.The radius of the deuteron orbit is approximately A)0.21 m B)1.8 m C)6.3 m D)10 cm E)27 cm

Question

Quizplus · Accepted Answer

The radius of the circular path can be found using the formula for the centripetal force:

F_c = qvB

where q is the charge of the deuteron, v is its velocity, and B is the magnetic field strength.

The centripetal force is also given by:

F_c = mv^2/r

where m is the mass of the deuteron and r is the radius of the circular path.

Equating these expressions for F_c, we get:

mv^2/r = qvB

Solving for r, we have:

r = mv/qB

Substituting the given values, we get:

r = (3.3 x 10^-27 kg) x (2.0 x 10^6 m/s) / (1.6 x 10^-19 C) x (0.40 T) = 0.10 m

Therefore, the radius of the deuteron's orbit is approximately 10 cm. The best choice is D.

A Deuteron Is Moving with a Speed of 2 ×\times× 106 M/s at Right Angles to a Magnetic Field

A Deuteron Is Moving with a Speed of 2 $\times$ 10⁶ M/s at Right Angles to a Magnetic Field