Consider a square with sides of length 5.0 cm I shown in the diagram. An electro
ID: 1550132 • Letter: C
Question
Consider a square with sides of length 5.0 cm I shown in the diagram. An electron is fixed at point P in the upper corner of the square. There is also a very long wire that runs along the bottom edge of the square carrying a current with a magnitude of 17 mA in the direction indicated by the diagram.A) what is the electric field at point Q (give both magnitude and direction).
B) what is the magnetic field at point Q (give both magnitude and direction).
C) suppose a particle with charge 4.9 pC passes through Point Q with a velocity of 5.2 m/s pointed at an angle of 30° above the +X axis. What is the net force acting on this particle. (give both magnitude and direction) consider a square with sides oflength s 0 cm, as hawn inthe diagram. An electron is fixed at point in the upper comer ofthe square, There is also a very long that along the bottom edge of the square carrying a current with a magnitude of 17 mA in the direction indicated by the diagram. A. What is the electric field point 0? Give both magnitude and di at B. what US both magnitude and direction.] is the magnetic field at point g suppose a particle with charge 49 uc passes through point o with a velocity of 5.2 m/s pointed at an angle of 30 above the *xaxis. What is the net force acting on this particle? [Give both magnitude and direction.] 2 m2. 30" t x axis. sin t 30 av B cine uge sin 12. 01 S fret
Explanation / Answer
(A)
Electric field at point Q,
E = k*q / d^2
E = 9*10^9*1.6*10^(-19) / (0.05)^2
E = 5.76*10^(-7) N/C
direction is along +x axis.
(B)
magnetic field at point Q,
B = u0*I / 2*pi* a
B = 4*pi*10^(-7)*17*10^(-3) / 2*pi*(0.05)
B = 6.8*10^(-8) T
Direction is into the page (-k).
(C)
v = 5.2 *(cos30 i + sin30 j) = 4.50i + 2.6j m/s
Electric force is,
Fe = q*E
F = 4.9*10^(-12)*5.76*10^(-7)
F = 2.825*10^(-18) iN
magnetic field,
Fm = q*(v x B)
Fm = 4.9*10^(-12)*[(4.50i + 2.6j) x (-6.8*10^(-8) k)]
Fm = 1.50*10^(-18)j - 0.867*10^(-18)i
Net force F = Fe + Fm
F = 1.958*10^(-18) i + 1.50*10^(-18) j
F = 2.46*10^(-18) N