An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

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Mary Herrera · Accepted Answer

a) Assume we have an electron accelerated using potential difference of V=350 V, which gives the ion a speed of v, to find this speed we set the potential energy equals to the kinetic energy of the ion:  eV=12mv2  v=2eVm  substitute with the givens to get:  v=2(1.6×10−19 C)(350 V)9.11×10−31 kg  =1.11×107 ms  b) Then the electron enters a region of uniform magnetic field, it moves in circular path with radius of:  r=mveB  substitute with givens to get:  r=(9.11×10−31 kg)(1.11×107 ms)(1.6×10−19 C)(200×10−3 T)  =3.16×10−4 m  r=3.16×10−4 m  Result: a) v=1.11×107 ms  b) r=3.16×10−4 m

An electron is accelerated from rest by a potential difference

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