Verified Solution • 2025–26 Session moderate

QJEEM20Ph55d3: JEE Main 2020 Physics - Class 12 JEEM20Ph55d3 Solution: A beam of protons with speed 4 × 10 5 ms –1 enters a uniform…

A beam of protons with speed 4 × 105 ms–1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to :
(Mass of the proton = 1.67 ×\times 10–27 kg, charge
of the proton = 1.69 ×\times 10–19 C)

Multiple Choice Options

  • A 2 cm
  • B 12 cm
  • C 5 cm
  • D 4 cm

Step-by-Step Verified Solution

Pitch = 2πmqB\frac{2\pi m}{qB} vcos θ\theta

= 2(3.14)(1.67×10−27)×4×105×cos⁡60(1.69×10−19)(0.3){{2\left( {3.14} \right)\left( {1.67 \times {{10}^{ - 27}}} \right) \times 4 \times {{10}^5} \times \cos 60} \over {\left( {1.69 \times {{10}^{ - 19}}} \right)\left( {0.3} \right)}}

= 0.04 m = 4 cm

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring