1. A long, straight wire is carrying a current of 8.0 A to the left. An electron
ID: 1886227 • Letter: 1
Question
1. A long, straight wire is carrying a current of 8.0 A to the left. An electron with charge 1.602 × 10-19 C is traveling to the right with a velocity of 1.2 m s1, parallel to the wire, a distance of 5.0 cm away from it, 5cm as per the diagram to the right. a. What is the strength of the magnetic field due to the current in the wire, at the position of 8.0A the electron? b. Is the magnetic field at the electron directed into or out of the page? c. What is the strength of the force on the electron due to the magnetic field from the wire? In what direction is this force? d. [4 marks]Explanation / Answer
a)
I = 8.0 A
r = distance at which magnetic field needs to be calculated
= 5 cm
= 0.05 m
B = miuo*I/(2*pi*r)
= (4*pi*10^-7)*(8.0) / (2*pi*0.05)
= (2*10^-7)*(8.0) / (0.05)
= 3.2*10^-5 T
Answer: 3.2*10^-5 T
b)
point the thumb of right hand in direction of current.
Then fingers will be give the direction of magnetic field
So, direction here is into the page
c)
|F| = |q|*v*B
= (1.602*10^-19)*(1.2)*(3.2*10^-5)
= 6.2*10^-24 N
Answer: 6.2*10^-24 N
d)
direction of B : into the page
direction of v : to the right
vXB will be up
since q is negative,
q(vXB) will be down
Answer: down
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