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When a charged particle moves at an angle of 11° with respect to a magnetic field, it experiences a magnetic force of magnitude f. at what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 1.6f? °?

Respuesta :

nurmaton
1) F=q[vB], where q -> charge, v-> velocity, B-> magnetic field. [ , ] -> cross product.
2) f=q*v*B*sin(11°) {1}
3) 1.6f=q*v*B*sin(alpha) {2}
4) {2} / {1} -> 1.6=sin(alpha)/sin(11°) or sin(alpha)=1.6*sin(11°) --> alpha=arcsin (1.6*sin(11°))
So, alpha=17.7°=18°

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