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(a) A charged particle having mass m and charge q is accelerated by a potential difference V, it flies through a uniform transverse magnetic field B. The field occupies a region of space d. Find the time interval for which it remains inside the magnetic field. (b) An alpha-particle is acceleration by a potential difference of 10^(4) V. Find the change in its direction of motion if it enters normally in a region of thickness 0.1 m having transverse magnetic induction of 0.1 T. (m_(alpha) = 6.4 xx 10^(-27) kg). ( c) A 10 g bullet having a charge of 4 muC is fired at speed of 270 m//sec in a horizontal direction. A vertical magnetic field of 500 muT exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations. |
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Answer» Solution :(a) `K = (1)/(2)MV^(2) - qV` `v = sqrt((2qV)/(m))` `sin theta = (d)/(R ) = (d)/(R ) = (d)/(mv// BQ) = (Bqd)/(mv)` (b) `V = 10^(4)V, q = 2e = 2 xx 1.6 xx 10^(-19)C`, `d = 0.1m, B = 0.1 T, m_(alpha) = 6.4 xx 10^(-27) kg` `K = (1)/(2)m_(alpha)v^(2) = qV = 2 eV` `v = sqrt((4 eV)/(m_(alpha)))` `R = (m_(alpha)v)/(Bq) = (m_(alpha)v)/(2eB)` `sin theta = (d)/(R ) = (d)/(m_(alpha)v//B.2e) = (2eBd)/(m_(alpha)v)` `= (2eBd)/(m_(alpha)sqrt((4eV)/(m_(alpha)))) = sqrt((E)/(m_alphaV)).Bd` `= sqrt((1.6 xx 10^(-19))/(6.4 xx 10^(-27) xx 10)) xx 0.1 xx 0.1` `= (1)/(2)` `theta = 30^(@)` ( c) `m = 10 g = 10 xx 10^(3) = 10^(-2) kg` `q = 4 muC = 4 xx 10^(-6)C` `v = 270 m//sec` `B = 500 muT = 500 xx 10^(-6) = 5 xx 10^(-4)T` `d = 100 m` `R = (mv)/(Bq) = (10^(-2) xx 270)/(5 xx 10^(-4) xx 4 xx 10^(-6)) = 13.5 xx 10^(8) m` `sin theta = (d)/(R ) = (100)/(13.5 xx 10^(8)) = (10^(-6))/(13.5)` `theta` is very small. DEFLECTION `= (d^(2))/(2R)` (as proved in previous example) `= ((100)^(2))/(2 xx 13.5 xx 10^(8)) = 3.7 xx 10^(-6) m` |
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