A 5.3×10 9 kg particle carrying acharge of 3.2×10 6 C is accelerated bya potenti

Question

A 5.3×109 kg particle carrying acharge of 3.2×106 C is accelerated bya potential difference of 3,200 V from rest. It then passesinto a uniform magnetic field of strength4.5×103 T.
(a) What is the largest force it can experience?
(b) What is the smallest force it can experience? A 5.3×109 kg particle carrying acharge of 3.2×106 C is accelerated bya potential difference of 3,200 V from rest. It then passesinto a uniform magnetic field of strength4.5×103 T.
(a) What is the largest force it can experience?
(b) What is the smallest force it can experience?

Explanation / Answer

We know that            F= qvBsin ...........1 According to conservation of energy we have K + P = 0 0.5mv2 - qV = 0    ==> v = 2qV /m            ..........2 Froom 1 and 2 we get            F= q(2qV / m)Bsin (a ) Largest force it can experience when =90o                F = q(2qV / m)Bsin90o                    =3.2×106 C*(2*3.2×106 C*3,200 V /5.3×109 kg)* 4.5×103 T*sin90o                    =............ N (b) Smallest force it can experience when =0o                         F= q(2qV / m)Bsin0o                                = 0 N                                = 0 N

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