A particle with charge ? 5.80 nC is moving in a uniform magnetic field B ? =?( 1

Question

A particle with charge ? 5.80 nC is moving in a uniform magnetic field B? =?( 1.20 T )k^. The magnetic force on the particle is measured to be F? =?( 4.00×10?7 N )i^+( 7.60×10?7 N )j^.

a- Are there components of the velocity that are not determined by the measurement of the force?

b- Calculate the x-component of the velocity of the particle.

c- Calculate the y-component of the velocity of the particle.

d-Calculate the scalar product v? ?F? .

e- What is the angle between v?  and F? ? Give your answer in degrees.

Explanation / Answer

here,

charge , q = - 5.8 nC = - 5.8 * 10^-9 C

magnetic feild , B = - 1.2 T k

F = (- 4 i + 7.6 j) * 10^-7 N

a)

Yes, the component of velocity parallel to the magnetic feild cannot be determined

b)

F = q * ( v X B)

(- 4 i + 7.6 j) * 10^-7 = - 5.8 * 10^-9 * ( v X ( - 1.2 k))

(- 57.4 i + 109.2 j) = ( v X ( 1 k))

solving for v

v = ( - 109.2 i - 57.4 j) m/s

the x-component of velocity is - 109.2 m/s

c)

the y-component of velocity is - 57.4 m/s

d)

v.F = ( - 109.2 i - 57.4 j) .((- 4 i + 7.6 j) * 10^-7 )

v.F = ( 4.37 - 4.36 ) * 10^-5 = 1 * 10^-7 N.m/s

e)

the angle between v and F , theta = arccos((v.F)/|v|.|F|)

theta = arccos((1 * 10^-7)/( 123.4 * 8.59 * 10^-7)) = 89.9 degree

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