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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMotion analysis in electric & magnetic fieldMedium3 minQB

PhysicsMediumnumerical

Passage / Comprehension

A positive, singly ionized atom of mass number $A_{M}$ is accelerated from rest by the voltage 192 V. Thereafter, it enters a rectangular region of width $w$ with magnetic field $B_{0} = 0.1 \hat{k}$ Tesla, as shown in the figure. The ion finally hits a detector at the distance x below its starting trajectory. [Given: Mass of neutron/proton $= (5/3) \times 1 0^{- 27} kg$ , charge of the electron $= 1.6 \times 1 0^{- 19} C$ .]\\

Find the value of distance x (in cm) for an ion with mass number $A_{m} = 144$ .

Answer:

48.00

Solution:

$v = 2 e V / m$ , $m = 144 \times (5/3) \times 1 0^{- 27} = 2.4 \times 1 0^{- 25} kg \Rightarrow v = 1.6 \times 1 0^{4} m / s$ .

$r = m v / e B_{0} = (2.4 \times 1 0^{- 25} \times 1.6 \times 1 0^{4}) / (1.6 \times 1 0^{- 19} \times 0.1) = 0.24 m$ .

The ion turns through a semicircle, so $x = 2 r = 0.48 m = 48 cm$ .

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Motion analysis in electric & magnetic field

⏱ 3mℹ️ Source: QB

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