Answer:
B = 5.59x10⁹ T
Explanation:
The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:
[tex] F = qvBsin(\theta) [/tex]
We have:
F = 1.4x10⁻³ N
v = 2.6x10⁶ m/s
θ = 37.0°
q = 2*p = 2*1.6x10⁻¹⁹ C
Hence, the strength of the magnetic field is:
[tex] B = \frac{F}{qvsin(\theta)} = \frac{1.4 \cdot 10^{-3}}{1.6 \cdot 10^{-19} C*2.6 \cdot 10^{6}*sin(37)} = 5.59 \cdot 10^{9} T [/tex]
Therefore, the strength of the magnetic field is 5.59x10⁹ T.
I hope it helps you!
