SA STATE Department of Chemistry 1. An electron traveling at 4.0x10 m\'s horizon
ID: 1865237 • Letter: S
Question
SA STATE Department of Chemistry 1. An electron traveling at 4.0x10 m's horizontally toward west in a vertically downward magnetic field of strength 1.5 T will experience a force of: A) zero N B)9.6x10" N upward C)9.6 x10 N towards east D) 9.6 x10 N, towards north E)9.6 x10N, towards south 2. In problem 1, the electron will rotate on a circle of radius A) zero m (does not circulate B)3.8 x10m C) 1.5 x10 m D) 4.6 x10 E) 1.7 x10 m 3. A particle of charge Q.50C and mass 1.0mg traveling eastward, enters into a region of uniform magnet field B, goes through half a sircle of radius 4. 0mm, and exit travelling westward. The speed of the particle as it enters/exits the region is 2.0m's (uniform circular motion). As the particle exit the A) F is castward and B is upward B) F is westward and B is downward C) F is northward and B is castward D) F is southward and B is westward E) F is downward and B is northward F) F is upward and B is southward 4. In problems 3, the magnitudes of the centripetal foree and magnetie field are respectively: A) zero N and zero T B) 0.5 xlo' N and 2.0 x10'T C) 1.0 xi0 N and 1.0x/0'T D) 2.0 x10 N and 0.5 x10'T E) 4.0 x10 N and 8.0 x/0 T 5. A straight wire carryiag an upward current I, 10.0 A. The magnetic field at point P (shown in the diagram), located at 0.020 m to the right (east) of the wire is: A) 1.0 xI0 T, towards north B) 1.0 x10, T, towards west C) 1.0 x10 'T, towards south D) 1.0x10'T, downward E) 1.0x/0'T, circulates clockwise 0.020m P 6. In problem 5, a second 1.00m-long wire carrying an upward current 1- 20.0 A is placed at point P shown above. The force on the second wire (due to the 1t wire) is: A) zero N B) 2.0x10 , N, towards west C)5.0 xi0'N, towards south E) 5.0 xIo N, downwards 7, A 0.40 m diameter loop of wire is initially oriented perpendicular to a 0.50 Tmagnetic field. The loon is D)2.0x/0'N, towards north so that its plane is parallel to the field direction in 0.20 second. The area of the loop is: A) 0.40 m B)0.16 ? C)0.13 m D1004 ? E) 0.02 mExplanation / Answer
here,
1)
speed of electron , v = 4 * 10^6 m/s to west
magnetic feild , B = 1.5 T downwards
the magnitude of force , F = q * v * B
F = 1.6 * 10^-19 * 4 * 10^6 * 1.5 N
F = 9.6 * 10^-13 N
using right hand rule
the direction of magnetic force is to the EAST
so, the correct option is C) 9.6 * 10^-13 N towards East
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