Exercise 27.8 Part B alculate the x-component of the velocity of the particle A

Question

Exercise 27.8 Part B alculate the x-component of the velocity of the particle A particle with charge 5.30 nC is moving ina uniform magnetic field B1.23 T)k. The magnetic force on the particle is measured to be F(3.40 10-7 N)i +(760 10-7 N) m/s Submit My Answers Give Up Part C Calculate the y-component of the velocity of the particle m/s Submit My Answers Give Up Part D Calculate the scalar product u·F m/s N Submit My Answers Give Up Part E What is the angle between v and F? Give your answer in degrees Submit My Answers Give Up Continue

Explanation / Answer

here,

q = -5.3 * 10^-9 C

B = - 1.23 T k

F = ( - 3.4 i + 7.6 j) * 10^-7 N

let the velocity be v

as F = q * ( v X B)

( - 3.4 i + 7.6 j) * 10^-7 = - 5.3 * 10^-9 * ( (vx i + vy j) X ( -1.23 k))

( - 64.15 i + 143 j) = ( (vx i + vy j) X ( 1.23 k))

( - 52.15 i + 116.5 j) = ( (vx i + vy j) X ( k))

( - 52.15 i + 116.5 j) = ( - vx j + vy i)

so on compairing

vx = - 116.5 m/s

vy = - 52.15 m/s

a)

the x-component of velocity is - 116.5 m/s

b)

the y-component of velocity is - 52.15 m/s

c)

v.F = ( - 116.5 i - 52.15 j) .( ( - 3.4 i + 7.6 j) * 10^-7) = - 2.4 * 10^-8 N.m/s

d)

let the angle be theta

cos(theta) = v.F /( |v| * |F|) = - 2.4 * 10^-8 /( 127.6 * 8.33 * 10^-7)

cos(theta) = -2.26 * 10^-4

theta = 90.01 degree

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