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QJEEM20Pha778: JEE Main 2020 Physics - Class 12 JEEM20Pha778 Solution: Photon with kinetic energy of 1MeV moves from south to north.…

Photon with kinetic energy of 1MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is 1.6 × 10–27 kg) :

Multiple Choice Options

  • A 0.71mT
  • B 7.1mT
  • C 0.071mT
  • D 71mT

Step-by-Step Verified Solution

JEE Main 2020 (Online) 8th January Morning Slot Physics - Magnetic Effect of Current Question 143 English Explanation

K.E = 12mv2

⇒ 1 MeV = 12mv2

⇒ 1.6 × 10-19 × 106 = 12 × 1.6 × 10–27 × v2

⇒ v = 2×107 m/s

Fm = qvB sinθ

⇒ Fm = qvB [as θ = 90o]

⇒ ma = Bev

⇒ 1.6 × 10–27 × 1012 = 1.6 × 10-19 × 2×107 × B

⇒ B = 0.71 mJ

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