5. east 6. north 7. There is no force. 009 (part 1 of4) 10.0 points The strength

Question

5. east 6. north 7. There is no force. 009 (part 1 of4) 10.0 points The strength of the Earth's magnetic field 012 (part 4 of 4) 10.0 points at the equator is approximately equal to If you are driving east, how fast would you 5 10-0 T. The force on a charge q moving have to drive in order for the magnetic force in a direction perpendicular to a magnetic on your head to equal 190 N (probably enough field is given by F qvB, where v is theto knock you over)? speed of the particle. The direction of the force is given by the right-hand rule. Suppose you rub a balloon in your hair and your head acquires a static charge of 2 × 10" . Answer in units of m/s If you are at the equator and driving west at a speed of 30 m/s, what is the strength of the magnetic force on your head due to the Earth's magnetic field? Answer in units of N 010 (part 2 of 4) 10.0 points What is the direction of that magnetic force? 1. upward 2. west 3. east 4. downward 5. south 6. north 011 (part 3 of 4) 10.0 points If you are at the equator and driving north at a speed of 30 m/s, what is direction of the magnetic force on your head? 1. south 2. downward 3. upwarod 4. west

Explanation / Answer

1)

Given :-

q = 2 x 10^-9 C

v = 30 m/s

B = 5 x 10^-5 T

F = qvB

F = 2 x 10^-9 C x 30 m/s x 5 x 10^-5 T

F = 3 x 10^-12 N

2)

Ans :- Downward

Applying the right-hand rule, the directionof the magnetic force is downward, since theEarth’s magnetic field points

toward magneticNorth, and the velocity is Westward.

3)

Ans :- There is no force

The force is zero when your velocity is par-allel to the magnetic field.

4)

F = 190 N

q = 2 x 10^-9 C

B = 5 x 10^-5 T

F = qvB

v = F / qB

v = 190 N / (2 x 10^-9 C x 5 x 10^-5 T)

v = 1.9 x 10^15 m/s

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