JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Magnetic Effects of Current

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A singly ionized magnesium atom (A24) ion is accelerated to kinetic energy 5 keV and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be ____ cm.

Answer: 10

Solution

The radius of the circular path in a magnetic field is

R = \frac{m v}{q B} = \frac{2 m K}{q B}

, where

K

is the kinetic energy. For a singly ionized Mg-24 ion:

m = 24 \times 1.66 \times 1 0^{- 27}

kg,

q = 1.6 \times 1 0^{- 19}

K = 5 \times 1 0^{3} \times 1.6 \times 1 0^{- 19}

J, and

B = 0.5

T. Calculating:

R = \frac{2 \times 24 \times 1.66 \times 1 0 ^{- 27} \times 5 \times 1 0 ^{3} \times 1.6 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19} \times 0.5} \approx 0.1

m = 10 cm.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer

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