An electron has a kinetic energy of 2.11E-17 J. It moves on a circular path that

Question

An electron has a kinetic energy of 2.11E-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 5.37E-5 T. Determine the radius of the path.

A long, straight wire carries a current of 46.2 A. The magnetic field produced by this current at a certain point is 7.88E-5 T. How far is the point from the wire?

The two wires shown in the figure below carry currents of I = 4.48 A in opposite directions and are separated by d0 = 8.40 cm.

Calculate the net magnetic field at a point midway between the wires. Use the direction into the page as the positive direction and out of the page as the negative direction.

Calculate the net magnetic field at point P1 - that is, d1 = 8.65 cm to the right of the wire on the right.

Calculate the net magnetic field at point P2 - that is, d2 = 20.1 cm to the left of the wire on the left

Tries 0/10 An electron has a kinetic energy of 2.11E-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 5.37E-5 T. Determine the radius of the path. A long, straight wire carries a current of 46.2 A. The magnetic field produced by this current at a certain point is 7.88E-5 T. How far is the point from the wire? The two wires shown in the figure below carry currents of I = 4.48 A in opposite directions and are separated by d0 = 8.40 cm. Calculate the net magnetic field at a point midway between the wires. Use the direction into the page as the positive direction and out of the page as the negative direction. Calculate the net magnetic field at point P1 - that is, d1 = 8.65 cm to the right of the wire on the right. Calculate the net magnetic field at point P2 - that is, d2 = 20.1 cm to the left of the wire on the left

Explanation / Answer

1.

since K.E = 1/2 mv^2
2K.E = mv^2
2k.E/r = QVb
r = 2k.E/QVB

r = (2*2.11*10^-17) / 1.6*10^-19*5.37*10^-5 = 4.9*10^6

2.

B= u I/(2 pi R)

u= 4 pi E-7 T m^2/ A

then R= u I/(2 pi B)

R= 4 pi E-7 x 46.2 /( 2 pi 7.88 E-5 )

R=11.725cm

3.

I'll solve this assuming the left wire is up and the right wire is down. If this is not the case basically the magnitude will be the same but the direction will be opposite

In my case both fields will be into the page and will have the same magnitude

So B = 2*(?o*I/2?r) = (but ?o/2? = 2.0X10^-7)

so B = 2*(2.0x10^-7*4.48/(0.084/2) = 4.2666x10^-5T (IN)

b) Here the left wire creates a field into the page and the right wire is out

so the net field will be IN

B = ?o/2?*(I1/0.0865 - I2/(0.084 + 0.0865)) = 2.0x10^-7*(4.48/0.0865 - 4.48/(0.084 + 0.0865) =

= 5.1x10^-6 T IN

c) Now the net field will be OUT

B = ?o/2?*(I1/0.201 - I2/(0.084 + 0.201)) = 2.0x10^-7*(4.48/0.201 - 4.48/(0.084 + 0.201) =

= 1.31x10^-6 T OUT

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