## An ion having charge +6e is traveling horizontally to the left at 8.30km/s when it enters a magnetic field that is perpendicular to its velo

Question

An ion having charge +6e is traveling horizontally to the left at 8.30km/s when it enters a magnetic field that is perpendicular to its velocity and deflects it downward with an initial magnetic force of 7.04*10^15 N .

What is the magnitude of this field?

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1 year 2021-08-15T01:30:12+00:00 1 Answers 51 views 0

Magnetic field, B = $$8.83\times 10^{29}\ T$$

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, $$F=7.04\times 10^{15}\ N$$

The magnetic force acting on the ion is given by the formula as follows :

$$F=qvB\ \sin\theta$$

Here, $$\theta=90^{\circ}$$

$$F=qvB\\\\B=\dfrac{F}{qv}\\\\B=\dfrac{7.04\times 10^{15}}{6\times 1.6\times 10^{-19}\times 8300}\\\\B=8.83\times 10^{29}\ T$$

So, the magnitude of magnetic field is $$8.83\times 10^{29}\ T$$.