A 5 \(\mu\) positively charged particle is moving at 45º to the direction of magnetic field with 3 x 10\(^4\)m/s speed. If it experiences a force of 6N, what is the value of the flux density of the field?
\(B = \frac{F}{q \cdot v \cdot \sin \theta}\)
\(\theta = 45^\circ \implies \sin 45^\circ = \frac{\sqrt{2}}{2}\)
Substituting the values:
\(B = \frac{6}{(5 \times 10^{-6})(3 \times 10^4) \left(\frac{\sqrt{2}}{2}\right)}\)
Simplifying:
\(B = \frac{6}{(15 \times 10^{-2}) \left(\frac{\sqrt{2}}{2}\right)} \approx \frac{6}{0.1061} \approx 56.5 \, \text{T}\)
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