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kolbaska11 [484]

3 years ago

6

3) An electron moves in a circular orbit of radius 0.11 mm, counterclockwise as you look down at it, moving in a plane above and

parallel to the plane of your exam paper. The velocity of the electron is v = 3 × 106m/s. a) Find the magnetic field due to the electron at the center of the circle. b) Is this magnetic field going into the page at the center of the circle or coming towards your face at the center of the circle as you look down at the electron?

1 answer:

andrey2020 [161]3 years ago

4 0

Answer:

B = 0.15 T

Explanation:

To find the magnitude of the magnetic field you use the following formula:

$B=\frac{mv}{qr}$ ( 1 )

m: mass of the electron = 9.1*10^-31 kg

v: velocity of the electron = 3*10^6 m/s

q: charge = 1.6*10^-19

r: radius = 0.11mm = 0.11*10^-3 m

You replace the values of the parameters in the equation (1).

$B=\frac{(9.1*10^{-31}kg)(3*10^6\frac{m}{s})}{(1.6*10^{-19}C)(0.11*10^{-3}m)}=0.15T$

the magnetic field has a magnitude of 0.15 T

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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