± Determining the Velocity of a Charged Particle Part A A particle with a charge

Question

± Determining the Velocity of a Charged Particle Part A A particle with a charge of-5.30 nC is moving in a uniform magnetic field of B 1.25 T )k, The magnetic force on the particle is measured to be r_-( 3.40×10-7 N ) i + ( 7.60×10-7 N )3. Are there components of the velocity that cannot be determined by measuring the force? yes no Submit Hints My AnswersGive Up Review Part Part B Calculate the x component of the velocity of the particle Express your answer in meters per second to three significant figures. ? 2 m/s Submit Hints My Answers Give Up Review Part Part C

Explanation / Answer

here,
Charge on particle, q = -5.30 nC = -5.30*10^9 C
magnatic field, B = (0i + 0j -1.25 k) T
Magnatic force on this particle , F = ( 3.40*10^7 i + 7.60*10^7 j + 0 K) N

Part A:
Yes, Z component will be Zero since F is zero along Z axis

Part B:

From Lortez Eqn in Component's of Forces'
Fx/q = (v2 * b3) - (v3 * b2)
3.40*10^7 /( -5.30*10^9) = (v2 * -1.25) - (v3 * 0)

v2 = -(6.415*10^-17 / 1.25)

v2 = vy = -5.132*10^-17 m/s

Similalry,
Fy/q = (v3 * b1) - (v1 * b3)
(7.60*10^7/(-5.30*10^9)) = v3*0 - v1*-1.25

v1 = vx = -1.147*10^-16 m/s

Similarly,
Fz/q = (v1 * b2) - (v2 * b1)
Fz = 0
so,
Vz = 0

The Velocity Vecotor, v = ( -1.147*10^-16 i - -5.132*10^-17 j + 0 k ) m/s

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