naiquawhite
07.04.2020 •
Mathematics
A singly charged ion of 7Li (an isotope of lithium) has a mass of 1.16×10−26 kg . It is accelerated through a potential difference of 300 V and then enters a magnetic field with magnitude 0.742 T perpendicular to the path of the ion.What is the radius of the ion's path in the magnetic field?
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Ответ:
8.87mm
Step-by-step explanation:
If the singled charged ion accelerates through a magnetic field, the force acting on the ion is expressed as;
F = qvBsin(theta)
Since the magnetic field is perpendicular to the path of the iron, angle that the ion made with the field will be 90°
F = qvBsin90°
F= qvB... (1)
q is the charge on the ion
v is the velocity possessed by the charge
B is the magnetic Field
Also according to Newton's second law, the force experienced by the ion can also be expressed as;
F = ma = mv²/R ... (2)
Where a is the centripetal acceleration = v²/R
m is the mass of the ion
R is the radius
Equating both value of the force to get the radius;
qvB = mv²/R
qB = mv/R
R = mv/qB ...(3)
Given m = 1.16×10^-26kg
q = 1.6×10^-19C
B = 0.742T
v = ?
Before we can get the radius R, we need to get the velocity of the charge in the wire using the relationship.
1/2mv² = eV (since the ion possess kinetic energy and potential difference V)
From the equation, v² = 2eV/m
v = √2eV/m
v = √2×1.6×10^-19×300/1.16×10^-26
v = √8.28×10^9
v = 90971.77m/s
Substituting v = 90971.77m/s into equation 3 to get the radius R of the ion's path in the field will give;
R = mv/qB
R = 1.16×10^-26 × 90971.77/1.6×10^-19×0.742
R = 1.056×10^-21/1.19×10^-19
R = 0.00887m
R = 8.87mm
Ответ:
B
Step-by-step explanation:
Take it one term at a time