A doubly charged ion with speed $6.9 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ enters a uniform $0.80 - T$ magnetic field, traveling perpendicular to the field. Once in the field, it moves in a circular arc of radius $30 \mathrm {~cm}$. What is the mass of this ion? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$
A) $11 \times 10 ^ { - 27 } \mathrm {~kg}$
B) $8.2 \times 10 ^ { - 27 } \mathrm {~kg}$
C) $3.3 \times 10 ^ { - 27 } \mathrm {~kg}$
D) $6.7 \times 10 ^ { - 27 } \mathrm {~kg}$

Question

Quizplus · Accepted Answer

The correct answer is A because the mass of the ion can be calculated using the formula:$$m = \frac{qBr}{v}$$where:- m is the mass of the ion- q is the charge of the ion- B is the magnetic field strength- r is the radius of the circular arc- v is the speed of the ionPlugging in the given values, we get:$$m = \frac{(2e)(0.80 \mathrm{~T})(0.30 \mathrm{~m})}{6.9 	imes 10^6 \mathrm{~m/s}} = 1.1 	imes 10^{-27} \mathrm{~kg}$$

A Doubly Charged Ion with Speed 6.9×106 m/s6.9 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }6.9×106 m/s

A Doubly Charged Ion with Speed $6.9 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$