The picture shows a region of uniform magnetic field B = 0.5 T out of the paper.

Question

The picture shows a region of uniform magnetic field B = 0.5 T out of the paper. In the bottom left hand corner of the region it a velocity selector of unknown electric field E which has a wall gap to allow only certain particles to pass through. The gap is a distance d = 10.0 cm from a plate. A sample of radioactive cobult is emitting alpha particles(|q| = 2e and m = 6.6413 times 10-27 kg) into the velocity selector. What must be the direction and magnitude of the minimum electric field in the velocity selector if the alpha particles are not to hit the plate? Is the ion positive or negative? Explain.

Explanation / Answer

   Given:   Applied magnetic field = B = 0.5 T   Let , v be the velocity of the alpha particle = ?    charge of the alpha particle = q = 2e = 2 (1.6 x10-19)  C     Mass of the particle = m = 6.6413 x10-27  kg    Since , Mangetic field applied ,    Kine tic enery of the particle is caused by this field   hence,           K.E =  q v B    (since :K.E= 1/2mv2)                                            v   =  2 q B / m                          = (2 x 3.2 x10-19 x 0.5 / 6.6413 x10-27  )                         =   0.69 x 104  m/s      Now , the electric field required in the velocity                   sector for this velocity is given by the another formua ,                     v = E / B                   or E = v B                          = (0.69 x104)(0.5)                          = 0.345 x 104                           or 3450 V/m                                             Applied magnetic field = B = 0.5 T   Let , v be the velocity of the alpha particle = ?    charge of the alpha particle = q = 2e = 2 (1.6 x10-19)  C     Mass of the particle = m = 6.6413 x10-27  kg    Since , Mangetic field applied ,    Kine tic enery of the particle is caused by this field   hence,           K.E =  q v B    (since :K.E= 1/2mv2)                                            v   =  2 q B / m                          = (2 x 3.2 x10-19 x 0.5 / 6.6413 x10-27  )                         =   0.69 x 104  m/s      Now , the electric field required in the velocity                   sector for this velocity is given by the another formua ,                     v = E / B                   or E = v B                          = (0.69 x104)(0.5)                          = 0.345 x 104                           or 3450 V/m                                          

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