Question
A particle with mass 0.8 gram and charge .02 μC is moving in the x-y plane with a velocity v(t) given by the formula v(t) = (15000 i + 30000 j )m/s. It enters a region of space where a magnetic field B exists, given by B = (1.5 i + 2j) T. What is the force on the charge? (Hint: i, j and k are the unit vectors along x, y and z-axis respectively. Use the relationships i x i = j x j = k x k = 0, and i x j = k, etc.)
a) .0003 N along positive z
b) .0003 N along positive x
c) .0003 N along negative z.
What is its acceleration?
0.375 m/s2 along negative z
b) 0.375 m/s2 along positive x
c) 0.375 m/s2 along positive z
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution
Trending nowThis is a popular solution!
Step by stepSolved in 4 steps
Knowledge Booster
Similar questions
- A particle with positive charge q = 3.52 x 1018 C moves with a velocity v = (5î + 4ĵ – k) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving particle, taking B = (3î + 2ĵ + k) T and E = (3î - j - 2k) V/m. (Give your answers in N for each component.) Ey = Ey = F, = (b) What angle does the force vector make with the positive x-axis? (Give your answer in degrees counterclockwise from the +x-axis.) ° counterclockwise from the +x-axis (c) What If? For what vector electric field would the total force on the particle be zero? (Give your answers in V/m for each component.) E, = V/m E, = V/m E, = V/marrow_forwardCurrent I = 30 A flows in a straight wire. An ion with charge Q = 3.2e-19 C is moving toward the wire, at distant r = 0.10 m, with speed v = 5.00×106 m/s. The magnetic force on the ion is antiparallel to the current. What is the magnitude of the force? (in N) OA: 3.742×10¯ OB: 4.379x10- OC: 5.123x10- OD: 5.994x10- 17 17 17 17 Submit Answer Tries 0/12 OE: 7.013x10 17 OF: 8.205x10- 17 OG: 9.600x10" OH: 1.123×10 17 16 Send Feedback 05008HISSOarrow_forwardYou have placed electrons in a magnetic field. The electrons move in horizontal circles of radius 9.5 cm when their speed is 2.0×10^6 m/s. What is the magnitude and direction of the magnetic field that causes this motion?arrow_forward
- A particle with charge 9.49 x 10^-6 C moves at 1.86 x 10^6 m/s through a magnetic field of strength 2.95 T. The angle between the particle s velocity and the magnetic field direction is 48.3 degrees and the particle undergoes an acceleration of 17.4 m/s^2. What is the particle s mass? 2.23 kg 8.7 kg 15.7 kg 22.6 kgarrow_forwardProblem 2: = A particle with charge +2.0 C moves through a uniform magnetic field. At one instant the velocity vector is v (2î+4ĵ + 6k) m/sec and the magnetic force on the particle is F = (41-20ĵ+ 12k) N. The x and y components of the magnetic field are equal (Bx = By). Thus, the magnetic field is given by B = B×î + B×ĵ + B₂k. What are B× and B₂? Answer: Bx = -3 T and B₂ = −4 T.arrow_forwardThis time I1 = 9.01, I2 = 1.40 A, and the two wires are separated by 8.41 cm. Now consider the charge q = 5.09 x 10^-6 C, located a distance of 7.00 cm to the right of wire I2, moving to the right at speed v = 66.0 m/s. What is the magnitude of the total magnetic force on this charge? 2.21E-09 N 1.11E-09 N 5.27E-09 N 3.69E-09 Narrow_forward
- A charged particle is entering a squared region of space with a uniform magnetic field. The sides of the region are 6 m wide. The particle enters the region exactly in the middle of one of the sides in a direction perpendicular to it, as in the Figure below. The charge of the particle is q = 20.0μC, its mass is m = 6.0 × 10-¹6 kg, and the velocity of the particle is |v| = 5 × 10³ m/s. How strong is the magnetic field so that the particle escapes such region in a direction perpendicular to the one it entered? See Figure for more details. XX XXXXXX :XXXXXXX XX XXX:arrow_forwardThe diagram above shows segments of two long straight wires, carrying currents. This time I1 = 6.46, I2 = 2.03 A, and the two wires are separated by 1.10 cm. Now consider the charge q = 6.74 x 10^-6 C, located a distance of 6.31 cm to the right of wire I2, moving to the right at speed v = 43.2 m/s. What is the magnitude of the total magnetic force on this charge? 3.34E-09 N 2.00E-09 N 6.95E-09 N 4.17E-09 Narrow_forwardThis time I1 = 5.03, I2 = 1.44 A, and the two wires are separated by 9.70 cm. Now consider the charge q = 7.27 x 10^-6 C, located a distance of 4.27 cm to the right of wire I2, moving to the right at speed v = 24.6 m/s. What is the magnitude of the total magnetic force on this charge? 1.19E-09 N 9.52E-10 N 2.49E-09 N 1.99E-09 Narrow_forward
arrow_back_ios
arrow_forward_ios