Answer:
Explanation:
Given that,
Charge on an electron is
q=1.609×10^-19C
Speed of electron V=4×10^5 m/s in positive x direction
V = 4×10^5 •i m/s
At this point F=0N
When the electron is moving in positive y direction
V =4×10^5 •j m/s
F = 3.2×10^-13 N in positive z direction.
F = 3.2×10 ^-13 •k N
What is the direction and magnitude of the magnetic field B?
The force in a magnetic field is given as
F = q (V ×B)
The only time V×B is zero is when both V and B are in same direction
This shows that B is in the positive x direction or negative x direction
i.e B= Bx •i
Now applying this to the second condition
F= q(V×B)
3.2×10^-13 •k =1.609×10^-19 × (4×10^5•j ×Bx •i)
Note that, j×i =-k
3.2×10^-13 •k / 1.609×10^-19 = (4×10^5Bx •(j×i))
1988812.927 •k =- 4×10^5Bx•k
1988812.927/4×10^5 •k =-Bx •k
4.972 •k =-Bx •k
Then, Bx = -4.972 •k Telsa
Then, the magnitude of B is
B=4.972 T
And its direction is negative z direction.
Answer:
B= 4.99T in negative x direction
Explanation:
F= qvBsin∅
q: charge on particle= 1.602 × 10^-19
v: velocity of the particle= 4 × 10⁵
B: Magentic field=???
∅: angle between magnetic field and direction of motion of charge particle
When moving in positive x- direction, the force is zero. According to the equation, the force will be zero when angle between direction of motion and magnetic field is either zero or 180⁰.
Now, when moving in positive y-direction, the force is 3.2 ˣ 10^-13 in positive z direction. and ∅=90. From Flemming's left hand rule, we can deduce that magnetic field is in negative x direction
3.2 ˣ 10^-13= 1.602 × 10^-19 × 4 × 10⁵ × B × sin90
B= 4.99T
