A proton moving in a uniform magnetic field with v⃗ 1=1.16×106ı^m/s experiences force F⃗ 1=1.61×10−16k^N. A second proton with v⃗ 2=2.26×106ȷ^m/s experiences F⃗ 2=−3.86×10−16k^N in the same field. A. What is the magnitude of B⃗ ?B. What is the direction of B⃗ ? Give your answer as an angle measured ccw from the +x-axis

B = [(1.07 × 10⁻³î) + (8.67 × 10⁻⁴j)] T

Magnitude of B = (1.377 × 10⁻³)

Direction of B = 39° anticlockwise from the +x-axis but in the clockwise direction, the direction becomes (360° - 39°) = 321°

Explanation:

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i think the answer is a and b

i wish you luck