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A charged particle with charge-to-mass ratio q/m = 1.0 x 10^8 C/kg…
Question
A charged particle with charge-to-mass ratio q/m = 1.0 x 10^8 C/kg enters a region of uniform magnetic field 0.10 T (directed perpendicular to its velocity) moving at 2.0 x 10^6 m/s. The field region is a slab of width 0.050 m measured along the original direction of motion. Find the angle (in degrees) by which the particle's velocity direction is deflected as it crosses the slab.
✓ Verified answer: 14.477512checked by our engine — not a guess
Step-by-step solution
Radius of circular path r = mv/(qB) = v/((q/m)B) = 2e6/(1e8 x 0.10) = 0.20 m.
As the particle traverses the slab, the chord's horizontal extent equals the slab width d.
The deflection angle theta satisfies sin(theta) = d/r = 0.050/0.20 = 0.25.
theta = arcsin(0.25) = 14.4775 deg.
Final answer14.4775
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