A positive charge moving in the direction of the^+x-axis enters a magnetic field

Question

A positive charge moving in the direction of the^+x-axis enters a magnetic field. If the charge experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the A) -y-axis. B) -x-axis C) +y-axis. D) +z-axis. E) -z-axis. 2) What can you say about the total magnetic flux through a closed surface? A) It is proportional to the current passing through the surface. B) It is proportional to the electric charge enclosed in the surface. C) It is proportional to the magnetic field enclosed in the surface. D) It is proportional to the magnetic charge enclosed in the surface. E) It is zero. 3) Which of the statements below is the primary significance of the Hall Effect experiment? A) It showed for the first time that there were no magnetic charges. B) It showed for the first time that charge carriers in conductors have negative charge. C) It showed for the first rime that the magnetic force was perpendicular to velocity. D) It showed for the first time that moving charges created magnetic fields. E) All of the above. 4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 - T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field? A) 60degree B)90degree C) 35degree D) 30degree E) 25degree 5) A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in an external magnetic field of 0.70 T with the plane of the coil making an angle of 30degree with the magnetic field. What is the magnetic torque on the coil? A) 0.088N middot m B) 0.076 N middot m C) 0.15 N middot m D) 0.40 N middot m E) 0.29 N middot m

Explanation / Answer

1) By Lorenz law,

F= q(v x B)

-j = +(i x k )   => Thus B has direction +Z axis.

2) By Ampere’s law,

B.dl = 0 I

Thus B is proportional to I

3) Hall effect first time showed that charge carriers in the conductor have negative charge.

4)

F= ILBsin

0.18 = 2.0*0.60*0.30*sin

= sin^-1[0.18/(2.0*0.60*0.3)] = 30 deg

5)

= NIABsin = 200*4.0*3.14*0.01^2*0.70*sin30 = 0.088 N.m

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