5 a current flow towards the east in a magnetic field that is directed North. Wh

Question

5 a current flow towards the east in a magnetic field that is directed North. What direction is the force on the current?

south

east

vertically up

vertically down

cannot be determined

6 two plates of a parallel plate capacitor have a charge to +- 2.0 nC when the capacitor is 4.0 V what is the capacitance?

8.0 nF

.50 nF

.125 nF

2.0nF

7 The electric field in a parallel plate capacitor is 5000 V/m. If the separation of the plates is 2.0mm what is the voltage difference between the two plates?

10 v

10,000V

2500V

4.0*10^-4 V

2.5*10^-6 V

8 what is the magnitude of the magnetic field due to a long straight wire carrying a current of 3.0 At a distance of 0.20 cm from the wire?

3.0 x 10^-7T

9.4 x 10^-7T

3.0 x 10^-4T

9.4 x 10^-4T

9 an electron Is acted on by a uniform electric field with a magnitude of 20 N/C what is the magnitude of the electrons acceleration (You may neglect all other forces on the electrons?

3.3e-26

3.2e-18

1.9e9

3.5e12

20

10 the potential difference between the two ends of a conducting wire is 3.0 V and the current through the wire is 20A what is the resistance of the wire?

6.7

.15

60

1200

.0225

11 in the previous question how much power is being dissipated in the wire?

6.7

.15

50

1200

.0225

12 a proton with a speed of 4.00 x 10^7 m/s in a magnetic field with a magnitude of .30 T. If the proton’s velocity makes an angle of 60 degrees with respect to the direction of the magnetic field what is the magnitude of the proton’s acceleration?

1.66e-13

1.92e-13

1.83e18

1.15e15

9.96e14

13 what is the equivalent resistance of the following combination of resistors?

1.8

.55

6.0

3.7

2.2

14 an electron moves with speed 1.0 x 10^5 m/s in the plane of the page as shown. A 3.00 x 10^-5 T magnetic field is directed out the page as shown what is radius and direction of the circle that the electron moves in?

.019 m clockwise

.019 m counterclockwise

1.9e3 clockwise

1.9e3 counterclockwise

cannot be determined

15 a long wire carries a current of 500.0 A to the right as shown. A 2.00 m length of wire carrying 500.0 A of current to the right is located a distance of 1.0 cm below the long wire. What is the resulting force on the 2.00 m length of wire?

.010 n

.10n

.31 n

5.0n

10 n

16 four charged oarticles are positioned at the corners of a square. What is the distance of the resulting electric field at the center of the square?

Toward the lower left corner

Toward the lower right corner

Toward the upper left corner

Toward the upper right corner

17 if each of the charges in the previous question are located 1.41 m from the center of the square what is the magnitude of the resulting electric field at the center of the square?

7.2e4

3.6e4

1.8e4

2.7e4

9.0e3

18 what potential difference is required to accelerate an electron from rest to 3.0 x 10^7 m/s?

7.5e-13

.16

2600

4.7e6

19 a uniform external magnetic directed into the pages goes through a loop of wire shown. The magnetic field increases at a steady rate starting from 0 and reaching a maximum of 60.0 mT after 60 s. The loop of wire is in the plane of the paper and has an area .010 m^2. What is the emf and the direction of the current induced in the loop while the field increases?

1.0e-5v clockwise

6.0e-4v counterclockwise

1.0e-5v clockwise

6.0e-4vcounterclockwise

.010v counterclockwise

20 the electric field due to a point charge is 30.0 n/c the electric potential at the same location is 3.0v (relative to a potential of zero at infinity). How far is the location from the charge?

.011

.10

.30

10

90

Explanation / Answer

5.

Using the Fleming's left hand rule, the direction of force would be vertically up.

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6.

for a capacitor

Q = CV, so

2nC = C(4V)

or, C = 0.5 nF

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7.

V = Ed = (5000 V/m)(2X10-3m)

or, V = 10 V

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8.

B = µ0I/2d = (2X10-7T-m/A)(3 A)/(0.002 m)

or, B = 3X10-4T

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9.

a = eE/m = (1.6X10-19C)(20N/C)/(9.1X10-31kg)

or, a = 3.5X1012 m/s2

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10.

R = V/I = 3V/20A

or, R = 0.15

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11.

Power dissipated is P = VI = 3V(20A)

or, P = 60 W

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12.

F = qvXB = qvBsin

a = qvBsin/m = (1.6X10-19C)(4X107 m/s)(0.3T)(sin600)/(1.67X10-27 kg)

or, a = 9.96X1014 m/s2

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13.

no resistors given.

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14.

figure needed

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15.

figure needed

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16.

magnitude of charge of the charged particles needed.

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17.

magnitude of charge of the charged particles needed.

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18.

Potential difference required is V,

then eV = (1/2)mv2

or, V = (0.5)(9.10X10-31kg)(3X107m/s)2/(1.6X10-19C)

or, V = 2600 V

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19.

figure needed

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20.

(1/40)q/r2 = 30N/C

(1/40)q/r = 3 V

or, r = 3/30

or, r = 0.1 m

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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