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PhysicsMagnetismMagnetic ForceHard2 minPYQ_2023
PhysicsHardnumerical

A proton with a kinetic energy of2.0 eVmoves into a region of uniform magnetic field of magnitudeπ2×10-3 T. The angle between the direction of magnetic field and velocity of proton is60o. The pitch of the helical path taken by the proton is ____cm. (Take, mass of proton=1.6×10-27 kgand charge on proton=1.6×10-19 C).

Answer:
40.00
Solution:

The given data is

K.E=2 eV

B=π2×10-3 T

θ=60o

The pitch of the proton is given by

P=2πmqB×vcosθ   ...(i)

By using

 K=mv22v=2Km

Substituting the values in equation (i)

=2π×2mKE×12×21.6×10-19×π×10-3

=2×2×1.6×10-27×2×1.6×10-19×1031.6×10-19

=2×2×101=0.4 m

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic Force
2mℹ️ Source: PYQ_2023

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