PHYS-2212-100 Hwk-06 FA-2017 Due: Name: MAGNETIC FORCES &FIELDS;(1) 1. (a) A neg

Question

PHYS-2212-100 Hwk-06 FA-2017 Due: Name: MAGNETIC FORCES &FIELDS;(1) 1. (a) A negative ion (charge -3.2 x 10-19 C) experiences a zero force when moving vertically upward at a speed of 3 x 103 m's in a certain magnetic field. However, a horizontal force of 6.0 x 10 N, directed toward the north, acts on the ion when it moves horizontally at the same speed toward the west. Find the magnitude and direction of the magnetic field in the region. (b) An electron moves with a velocity of (21-J- 2k) x 10 m/s in a region of space where the magnetic field is given as 31 - 5) x 109 T. Find the magnitude of the magnetic force exerted on the electron. [(a) 0.625 T, vertically downward]; (b) 2.18 x 10"N 2. A 1.2-m length of rigid wire of mass 15 g carries an electric current of 2.0 A in the negative x-direction inside a region of space where the external magnetic field is given as 3t-5k x 102 T. (a) Find the magnetic force exerted on this wire. [(") 0.12 N] (b) Determine the current direction and the quantity of current just necessary to levitate or suspend this wire inside the given magnetic field. [2.1 A, positive x-direction A singly charged positive ion (mass = 3.2 x 1026 kg) is accelerated through a potential difference of 1000 V before the ion enters a uniform magnetic field in a direction perpendicular to the field. What strength of the magnetic field that would be required to constrain the ion to move in a circular orbit ofradius 5.0 cm? [0.4 T 3- 4. Direct electric current of 25 mA is maintained in a circular loop of radius 10 cm and consisting of 100 turns. An external magnetic field, 0.5 T, is directed at an angle of 60° with the plane of the loop. (a) Calculate the magnitude of the torque exerted on the loop by the magnetic field (b) What is the magnitude of the magnetic moment of the current loop? (c) How much work must be done to rotate the loop from its present orientation in the magnetic [1.96 x 10 N.m] [7.85 x 10 A.m) field to the position when its magnetic dipole moment is aligned with the magnetic field? [5.3 x 10" 5. Calculate the magnitude of the magnetic field at point Pz due to a semi infinitely-long wire bent at right angles as shown. The wire carries a steady current 1-10 A. 4 cm 4 cm 4 cm 3cm [4.8 x 10 T, out of page)

Explanation / Answer

1. (A) F = q ( v x B)

when v -> k^ then F = 0

so x and y component of magnetic field are zero.

when v = - 3 x 10^5 i

then F = 6 x 10^-14 j

(6 x 10^-14)j = (-3.2 x 10^-19)[(-3 x 10^5)i x By]

By = 0.625(- k )

Field strength is 0.625 T and downward.

(b) q = -1.6 x 10^-19

v = (2i - j - 2k) x 10^4

B = (3i - 5j) x 10^-3

F = q (v X B)

F = 1.6 x 10^-17 k - 4.8 x 10^18 k + 9.6 x 10^-18j + 1.6 x 10^-17 i

F = (16 x 10^-18)i + (9.6 x 10^-18)j + (11.2 x 10^-18)

magnitude of force = sqrt(16^2 + 9.6^2 + 11.2^2) x 10^-18

= 21.8 x 10^-18 N

= 2.18 x 10^-17 N

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