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?
A
107.23T
B
75.76T
C
98.44T
D
56.57T
Verified Correct Answer: D.
Physical Principles & Numerical Solution
\(B = \frac{F}{q \cdot v \cdot \sin \theta}\)
\(\theta = 45^\circ \implies \sin 45^\circ = \frac{\sqrt{2}}{2}\)
Substituting the values:
\(B = \frac{6}{(5 \times 10^{-6})(3 \times 10^4) \left(\frac{\sqrt{2}}{2}\right)}\)
Simplifying:
\(B = \frac{6}{(15 \times 10^{-2}) \left(\frac{\sqrt{2}}{2}\right)} \approx \frac{6}{0.1061} \approx 56.5 \, \text{T}\)
Source: Myschool Past Questions Archive (Record #74487)