Question

A proton enters a magnetic field with a velocity of 2.5 X 10^7m/s making angle 30
with the magnetic field. What is the force on the proton? Given that, B= 2.5T​

Answer 1

Answer:

FORCE = 5 × 10-¹² N [ 5 PN ]

Explanation:

★ MAGNETISM ★

★ Given ;

» Charge of Particle ( Q ) = 1.6 × 10^-19 C ( Proton )

» Velocity of Particle ( V ) = 2.5 × 10^7

m / s

» Magnetic Field ( B ) = 2.5 Tesla

» Angle made between Velocity Vector and Magnetic Field ( ∅ ) = 30°

★ FORCE ON PARTICLE ( F ) = ???

_____ [ BY USING FORMULA ] _____

$[ Force \: ( \: F \: ) \: = \: Charge \: of \: Particle \: (Q) \: \times \: Velocity \: Vector \: (V) \: \times \: Magnetic \: Field \: (B) \: \times \: \sin( \alpha ) ]$

★ F = Q × V × B sin∅

» F = 1.6 × 10^-19 × 2.5 × 10^7 × 2.5 × Sin∅

» F = 4 × 10-¹² × 2.5 × Sin 30°

» F = 10 × 10-¹² × 1 / 2

» F = 5 × 10-¹²

★ FORCE ( F ) = 5 × 10-¹² N

----------------------- [ OR ] ----------------------

★ FORCE ( F ) = 5 PICO NEWTON

_______________________________________

ANSWER :- FORCE ON PROTON IN MAGNETIC FIELD IS ;

[ F = 5 × 10-¹² N ]