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A charged particle of mass m and charge q is projected on a rough horizontally xy- plane surface with z- axis in the vertically upward direction. Both electric and magnetic fields are acting in the region and given by vec(E)=-E_(0)hat(k) and vec(B)=-B_(0)hat(k) respectively. The particle enters into the field at (a,0,0) with velocity vec(v)=upsilon_(0)hat(j). The particle starts moving into a circular path on the plane. If the coefficient of friction between the particle and the plane is mu. Then calculate tha : (a) time when the particle will come to rest (b) time when the particle will hit the centre. ( c) distance travelled by the particle when it comes to rest. |
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Answer» Solution :`(a) N=mg+qE_(0)……………….(1)` `qB_(0)upsilon=(m upsilon^(2))/( R )………………(2)` and `-m(d upsilon)/(DT) = mu N …………………(2)` From equation `(2), ""R=( m upsilon)/(B_(0)q)……………..(4)` From equations `(1)` and `(3)` `-m.(d upsilon)/(dt)=mu ( mg +qE_(0))( o r)` `-m int_(v_(0))^(0)d upsilon=mu(mg+qE_(0))int_(0)^(t)dt` Thus `t=(m upsilon_(0))/(mu(mg+qE_(0)))` `(b)`From equation `(4)` `dR=(m)/(B_(0)q)d upsilon=-(mu(mg+qE_(0))dt)/(qB_(0))` `int_(R_(1))^(0)dR=(-mu(mg+qE_(0)))/(qB_(0))INT _(0)^(t)dt` or or`t=(qB_(0)R_(i))/(mu(mg+qE_(0)))` Here `R_(i)=(m upsilon_(0))/(B_(0)q)` Thus`t=(m upsilon_(0))/(mu(mg+qE_(0)))` `(c )``-m upsilon.(d upsilon)/(dl)=mu (mg+qE_(0))` or, `-m int_(upsilon_(0))^(0)upsilond upsilon=mu(mg+qE_(0))int_(0)^(t)dt ` or `l=(m upsilon_(0)^(2))/(2mu(mg+qE_(0)))` |
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