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4 years ago

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A proton travels with a speed of 5.05 ✕ 106 m/s at an angle of 64° with the direction of a magnetic field of magnitude 0.160 T i

n the positive x-direction.(a) What is the magnitude of the magnetic force on the proton? N
(b) What is the proton's acceleration? m/s2

2 answers:

Natali [406]4 years ago

7 0

Answer:

(a) the magnitude of the magnetic force on the proton is 1.16 x 10⁻¹³ N

(b) the proton's acceleration is 6.97 x 10¹³ m/s²

Explanation:

given information:

proton's speed, v = 5.05 ✕ 10⁶ m/s

angle, θ = 64°

magnetic field, B = 0.160 T ( +x direction)

(a) the magnitude of the magnetic force on the proton

F = q v B sin θ

where

F = magnetic force (N)

q = charge (C)

v = speed (m/s)

B = magnetic field (T)

θ = angle

F = q v B sin θ

proton charge, q = 1.602 x 10⁻¹⁹ C

thus,

F = (1.602 x 10⁻¹⁹) (5.05 ✕ 10⁶) (0.160) sin 64°

= 1.16 x 10⁻¹³ N

the proton's acceleration

F = m a

where

m = mass (kg)

a = acceleration (m/s²)

the mass of proton, m = 1.67 x 10⁻²⁷ kg

so,

F = m a

a = F/m

= (1.16 x 10⁻¹³)/(1.67 x 10⁻²⁷)

= 6.97 x 10¹³ m/s²

Setler [38]4 years ago

4 0

Answer:

A. 1.19 * 10^(-13) N

B. 7.12 * 10^(15) m/s²

Explanation:

Parameters given:

Speed, v = 5.05 * 10^6 m/s

Angle, A = 64°

Magnetic field strength, B = 0.160T

Mass of proton, m = 1.673 * 10^(-27) kg

Charge of proton, q = 1.6023 * 10^(-19) C

A. Magnetic force is given as:

F = q*v*B*sinA

F = 1.6023 * 10^(-19) * 5.05 * 10^6 * 0.160 * sin64

F = 1.19 * 10^(-13) N

B. Force is generally given as:

F = m*a

Hence, we can find acceleration, a, by making it the subject of formula:

a = F/m

a = (1.19 * 10^(-13))/(1.673 * 10^-27)

a = 7.12 * 10^15 m/s²

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