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Serjik [45]
3 years ago
6

A singly charged ion of 7Li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. It is accelerate

d through a potential difference of 224 V and then enters a magnetic field with magnitude 0.724 T perpendicular to the path of the ion. What is the radius of the ion’s path in the magnetic field? (Give your answer in decimal using "mm"(millimeter) as unit)
Physics
1 answer:
MaRussiya [10]3 years ago
4 0

Explanation:

It is given that,

Mass of lithium, m=1.16\times 10^{-26}\ kg

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

\dfrac{1}{2}mv^2=qV

v=\sqrt{\dfrac{2qV}{m}}

q is the charge on an electron

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}

v = 78608.58 m/s

v=7.86\times 10^4\ m/s

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

qvB=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

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Answer: 1.88 N

Explanation:

Data:

Force = 4.00N
angle = 62°
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Solution:

The trigonometric ratio that relates horizontal - leg to hypotenuse is the cosine.

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