Answer:
Magnetic field, B = [tex]8.83\times 10^{29}\ T[/tex]
Explanation:
Given that,
Charge in the ion, q = 6e
Speed of the ion, v = 8.3 km/s = 8300 m/s
Magnetic force acting on the ion, [tex]F=7.04\times 10^{15}\ N[/tex]
The magnetic force acting on the ion is given by the formula as follows :
[tex]F=qvB\ \sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=qvB\\\\B=\dfrac{F}{qv}\\\\B=\dfrac{7.04\times 10^{15}}{6\times 1.6\times 10^{-19}\times 8300}\\\\B=8.83\times 10^{29}\ T[/tex]
So, the magnitude of magnetic field is [tex]8.83\times 10^{29}\ T[/tex].
