2 A proton (mp = 1.67×1027kg, qp = 1.602×1019C is fired down the y-axis in the +

Question

2 A proton (mp = 1.67×1027kg, qp = 1.602×1019C is fired down the y-axis in the +ˆy direction at a velocity ~v = 100m/sˆy. The coordinate system is chosen so that the force of gravity acts in the zˆ direction.

(a)What must the direction of a magnetic field be in order to create an upward force (+ˆz) on the proton? Justify.

(b)What magnitude must the magnetic field have so that the proton travels in a straight line?

(c)If the magnetic field is provided by a wire carrying a current, I, and the wire is 10cm from the path of the particle. How much current must be sent through the wire to generate the field in (b)?

Explanation / Answer

Solution:

a) The force is in the positive z direction The magnetic force is perpendicular to the velocity at any of time.

b) The proton experiences a force mg due to its own weight and a force due to magnetic field .

qvB = mg

B = mg / qv = ( 1.67 x 10^-27) ( 9.8) / (1.6 x 10^-19)(100) = 1.02 x 10 ^ -9 T

c) If the length L = 10 cm = 0.10 m

Magnetic field = B = 1.02 * 10^-9 T

Current = I

F= B I L

=>mv^2/L = B I L

=> I = mv^2 / BL^2 = (1.67 x 10^-27) (100)^2 / (1.02 x 10^-9)(0.10) = 1.6 x 10^-13 A

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.