Question

an Alpha particle experiences a force of 3.84*10^-14 N when it moves perpendicular to magnetic field of 0.2 wb/m^2 then speed of the alpha particle is:
a) 6*10^5 m/sec
b) 5*10^5 m/sec
c) 1.2*10^6 m/sec
d) 3.8*10^6 m/sec

Answer 1

Answer:

✅ VELOCITY OF ALPHA PARTICLE (V) = 6 × 10^5 m / s . ✅

▶ OPTION A ◀

Explanation:

★ MAGNETIC FORCE ★

★ Given ;

» Magnetic Force ( F ) = 3.84 × 10-¹⁴ N

» Magnetic Field ( B ) = 0.2 Weber / m²

» Charge of ALPHA PARTICLE ( q ) = 2 × q ( p ) [ charge of proton ]

» Angle of Magnetic Force with Velocity vector = Perpendicular ( 90° )

★ SPEED OF ALPHA PARTICLE ( V ) = ???

______ [ BY USING FORMULA ] _______

★ $Magnetic \: Force \: ( \: f \: ) \: = \: charge \: of \: Alpha \: Particle \: ( \: q \: ) \: \times \: Velocity \: ( \: v \: ) \: \times \: Magnetic \: field \: ( \: b \: ) \: \times \sin( \alpha )$ ★

★ F = q × V × B × sin∅ ★

» 3.84 × 10-¹⁴ = 2 × ( 1.6 × 10^-19 ) × V × 0.2 × Sin 90°

» 3.84 × 10-¹⁴ = 32 × 10^-20 × V × 0.2 × 1

» V = 3.84 × 10-¹⁴ / 32 × 0.2 × 10^-20

» V = 3.84 × 10-¹⁴ / 6.4 × 10^-20

» V = 0.6 × 10-¹⁴ × 10^20

» V = 0.6 × 10^6

★ » Velocity ( V ) = 6 × 10^5 m / s « ★

———————— [ OR ] ————————

⏩ Velocity ( V ) = 600 Km / Second ⏪

________________________________________

ANSWER :- VELOCITY OF ALPHA PARTICLE having both Magnetic Force ( F ) and Magnetic Field ( B ) is ;

♣ V = 6 × 10⁴ m / s ♥

Answer 2

Answer:

★ F = q × V × B × sin∅ ★

» 3.84 × 10-¹⁴ = 2 × ( 1.6 × 10^-19 ) × V × 0.2 × Sin 90°

» 3.84 × 10-¹⁴ = 32 × 10^-20 × V × 0.2 × 1

» V = 3.84 × 10-¹⁴ / 32 × 0.2 × 10^-20

» V = 3.84 × 10-¹⁴ / 6.4 × 10^-20

» V = 0.6 × 10-¹⁴ × 10^20

» V = 0.6 × 10^6

★ » Velocity ( V ) = 6 × 10^5 m / s « ★

———————— [ OR ] ————————

⏩ Velocity ( V ) = 600 Km / Second ⏪

________________________________________

Explanation: