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bixtya [17]
3 years ago
7

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

ion enters a magnetic field of 0.522 T in a direction perpendicular to the field. The charge on the ion is 1.602 × 10−19 C. Find the radius of the ion’s path in the field. Answer in units of cm.
Physics
1 answer:
jasenka [17]3 years ago
5 0

Answer:

1.8 cm

Explanation:

m = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

q = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

\Delta V =Potential difference through which the ion is accelerated = 215 V

v = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}

r = Radius of the path followed by ion

B = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm

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