Question

An alpha particle experiences a force of 3.84*10^-14 N when it's moves perpendicular to magnetic field of .2Wb/m^2 then the speed of alpha particle is

Answer 1

Answer:

$6.0\times 10^5\ m/s$.

Explanation:

Given:

• Force on alpha particle, $F = 3.84\times 10^{-14}\ N$.

• Strength of magnetic field, $B =  0.2\ Wb/m^2$

The alpha particle is moving perpendicular to the magnetic field therefore the angle between its velocity and magnetic field is $90^\circ$.

We know,

The charge on an alpha particle is $+2e$.

$e=\text{magnitude of charge on an electron}=1.6\times 10^{-19}\ C.$

Assumptions:

• $\theta$ = angle between magnetic field and velocity of the alpha particle = $90^\circ$.

• $q$ = charge on the alpha particle = $2e=2\times 10^{-19}\ C$.

• $v$ = velocity of the alpha particle.

The magnetic force on a charged particle moving in a magnetic field is given by

$\vec F = q(\vec v \times \vec B)\\\text{In terms of magnitude, }\\F=qvB\sin\theta\\3.84\times 10^{-14}=(2\times 1.6\times 10^{-19})\times v\times  (0.2)\\v=\dfrac{3.84\times 10^{-14}}{(2\times 1.6\times 10^{-19})\times(0.2)}=6.0\times 10^5\ m/s.$

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 ⏪

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Explanation: