A particle of mass `m = 1.6 xx 10^(27)` kg and charge `q = 1.6 xx 10^(-19)` C moves at a speed of `1.0 xx10^7 ms^(-7)`. It enters a region of uniform magnetic field at a point E, as shown in The field has a strength of 1.0 T. (a) The magnetic field is directed into the plane of the paper. The particle leaves the region of the filed at the point F. Find the distance EF and the angle theta. (b) If the field is coming out of the paper, find the time spent by the particle in the region of the magnetic field after entering it at `E`.

Explanation

(a) As th particle particle enters the magnetic field, it
will travel in a circular path.The centre will be on the
line perpendicularto its velocity and the redius r will
`be(mv)/(qB).` The direction of the force `vecqvX vecB` shows that the
centre will be outside the field as shown in.
As `leAEO = 90^@` (as AE is tangent and OE is
radius) and, `leAEC = 45^@,` we have `leOEF = 45^@.`
As OE = OF (they are radii of the circular arc),
`le OFE = le OEF = 45^@.` also, OF is perpendicular to the
velocity of the particule at F, so that `theta =45^@`. from
triangle OEF,
`EF= 2.OE cos leOEF`
`= 2.(mv)/ (qB). (1)/sqrt(2)`
= `(sqrt(2 xx (1.6 X10^(27)kg)xx (10^7 ms^(-1)/(1.6xx 10^(-19) C) xx1.0 T)`
`= sqrt(2xx10^(-1)m 14cm.`
If the magnetic field is coming out of the paper, the
direction of the force `vecqv xvec B` shown in figure Again `leAEO = 90^@,` giving
`leOEF= leOFE=45^@.` Thus, the angle `EOF = 90^@ `The particle describes three fourths of the complete circle inside the field. As the speed v is uniform, the time spent in the magnetic field will be `(3)/(4) X (2pir)(v) = (3pimv)(2vqB) = (3pim)/(2qB)`
`= (3X3.14X1.6X10^(-27) kg)/(2X1.6 X 10^(-19) CX1.0 T) =4.7 X 10^(-8) s.`