A particle with charge 6.00 nC is moving in a uniform magnetic field B =( 1.21 T

Question

A particle with charge 6.00 nC is moving in a uniform magnetic field B =( 1.21 T )k^. The magnetic force on the particle is measured to be F =( 3.90×107 N )i^+( 7.60×107 N )j^

Part B

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

Part C

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

Part D

Calculate the scalar product v F

Part E

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

Part D

Calculate the scalar product v F

Part E

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

Explanation / Answer

here,

charge , q = - 6 nC = - 6 * 10^-9 C

magnetic feild , B = - 1.21 T k

magnetic force , F = ( 3.9 i + 7.6 j) * 10^-7 N

let the velocity be v

F = q *( v X B)

( 3.9 i + 7.6 j) * 10^-7 = - 6 * 10^-9 * ( v X( 1.21 k))

( 3.9 i + 7.6 j) * 16.67 = - 1 * ( v X( 1.21 k))

so v = (104.7 i - 53.7 j) m/s

B)

the x-component of the velocity of the particle is 104.7 m/s

C)

the y-component of the velocity of the particle is - 53.7 m/s

D)

v .F = ( 104.7 i - 53.7 j) .( ( 3.9 i + 7.6 j) * 10^-7) = 2.1 * 10^-8 N.m/s

E)

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

theta = arccos( 2.1 * 10^-8 /( 106.67 * 8.54 * 10^-7))

theta = 89.986 degree

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