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Magnetics question

2022 · 28 Jun · Shift 1 · Q68
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  5. /2022 · 28 Jun · Shift 1 · Q68

Magnetics question

2022 · 28 Jun · Shift 1 · Q68

JEE MainPhysicsMagneticsNumerical+4 / −1
A singly ionized magnesium atom (A = 24) 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 ‾\underline{\hspace{2cm}}​ cm.
Numerical answer
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Correct answer: 10

  1. Given data
  • Singly ionized magnesium ion: Mg+\mathrm{Mg}^+Mg+
  • Mass number: A=24A=24A=24
  • Kinetic energy: K=5 keV=5×103×1.6×10−19 JK=5\,\text{keV}=5\times 10^3\times 1.6\times 10^{-19}\,\text{J}K=5keV=5×103×1.6×10−19J
  • Magnetic field: B=0.5 TB=0.5\,\text{T}B=0.5T
  • Since it is singly ionized, charge magnitude is q=e=1.6×10−19 Cq=e=1.6\times 10^{-19}\,\text{C}q=e=1.6×10−19C
  1. Mass of magnesium ion

For A=24A=24A=24, the ion mass is approximately m=24u=24×1.66×10−27 kgm=24u=24\times 1.66\times 10^{-27}\,\text{kg}m=24u=24×1.66×10−27kg m=3.984×10−26 kgm=3.984\times 10^{-26}\,\text{kg}m=3.984×10−26kg

  1. Use kinetic energy to find speed

We know K=12mv2K=\frac{1}{2}mv^2K=21​mv2 So, v=2Kmv=\sqrt{\frac{2K}{m}}v=m2K​​

Now, K=5×103×1.6×10−19=8×10−16 JK=5\times 10^3\times 1.6\times 10^{-19}=8\times 10^{-16}\,\text{J}K=5×103×1.6×10−19=8×10−16J

Thus, v=2×8×10−163.984×10−26v=\sqrt{\frac{2\times 8\times 10^{-16}}{3.984\times 10^{-26}}}v=3.984×10−262×8×10−16​​ v=1.6×10−153.984×10−26v=\sqrt{\frac{1.6\times 10^{-15}}{3.984\times 10^{-26}}}v=3.984×10−261.6×10−15​​ v=4.016×1010v=\sqrt{4.016\times 10^{10}}v=4.016×1010​ v≈2.0×105 m/sv\approx 2.0\times 10^5\,\text{m/s}v≈2.0×105m/s

  1. Radius of circular path in magnetic field

Since the ion enters perpendicular to B⃗\vec BB, the radius is r=mvqBr=\frac{mv}{qB}r=qBmv​

Substitute values: r=(3.984×10−26)(2.0×105)(1.6×10−19)(0.5)r=\frac{(3.984\times 10^{-26})(2.0\times 10^5)}{(1.6\times 10^{-19})(0.5)}r=(1.6×10−19)(0.5)(3.984×10−26)(2.0×105)​

First numerator: 3.984×10−26×2.0×105=7.968×10−213.984\times 10^{-26}\times 2.0\times 10^5=7.968\times 10^{-21}3.984×10−26×2.0×105=7.968×10−21

Denominator: 1.6×10−19×0.5=8.0×10−201.6\times 10^{-19}\times 0.5=8.0\times 10^{-20}1.6×10−19×0.5=8.0×10−20

Therefore, r=7.968×10−218.0×10−20=0.0996 mr=\frac{7.968\times 10^{-21}}{8.0\times 10^{-20}}=0.0996\,\text{m}r=8.0×10−207.968×10−21​=0.0996m

  1. Convert to cm

0.0996 m=9.96 cm≈10 cm0.0996\,\text{m}=9.96\,\text{cm}\approx 10\,\text{cm}0.0996m=9.96cm≈10cm

  1. Final answer

The radius of the path is 10 cm\boxed{10\,\text{cm}}10cm​

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