PHYSICS 1E03 Lecture Notes - Lecture 1: Uranium-238, Electronvolt

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5 May 2016

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Answers: v = 1. 98x10(4) m/s (positive x-direction) a = 2. 03x10(12) m/s^2 (positive z direction) E = 19. 7 n/c (positive z direction) F = m a = (9. 11x10(-31)) (2. 03x10(12)) = 1. 85x10(-18) n. F = q v b + q e. Therefore: the magnetic field on the y-component is -1. 58 x 10^-3 t, the magnetic field on the z-component is 0 t, length of cube"s edge = 38. 6 cm = . 386m. B = 0. 0198 t: the magnitude of the magnetic force on segment ab, the magnitude of the magnetic force on segment bc, the magnitude of the magnetic force on segment cd: The y-component of the following will be cancelled as the direction of the current is parallel to b, so theta = 0. So only, take into account of the x-component = part (b): the magnitude of the magnetic force on segment da: The sqrt(2) results because of the sqrt(x + y component) that makes the diagonal wire across the cube.

Related Questions

The cube is 40.0 cm on each edge. Four straight segments ofwireâab, bc, cd, and daâform a closed loop that carries a current I= 5.00 A . A uniform magnetic field of magnitude B = 0.020 0 T isin the positive y-direction. Determine the magnitude and directionof the magnetic force on each segment.

A rectangular loop of wire with sides H = 30 cm and W = 67 cm is located in a region containing a constant magnetic field B = 1.11 T that is aligned with the positive y-axis as shown. The loop carries current I = 259 mA. The plane of the loop is inclined at an angle with respect to the x-axis.

1)What is , the x-component of the magnetic moment vector of the loop?

2)What is , the y-component of the magnetic moment vector of the loop?

3)What is , the z-component of the torque exerted on the loop?

4)What is F_bc, the magnitude of the force exerted on segment bc of the loop?.

5)What is the direction of the force that is exerted on the segment bc of the loop?

along negative x-direction
along negative y-direction
along positive x-direction
along positive y-direction

none of the above

copperwalrus365

An alpha particle (q = +3.20×10^-19 C) moves at a velocity of 2.45×10^4 m/s in the positive x direction through a magnetic field of 0.0725 T that points in the negative z direction. What are (a) the magnitude in N and (b) the direction of the magnetic force on the particle?

PHYSICS 1E03 Lecture Notes - Lecture 1: Uranium-238, Electronvolt

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