An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall
An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall
- A
$B = \frac{{2bm{v_0}}}{{\left( {{b^2} - {a^2}} \right)e}}$
- B
$B = \frac{{2bm{v_0}}}{{\left( {{a^2} + {b^2}} \right)e}}$
- C
$B = \frac{{m{v_0}}}{{\left( {\sqrt {{b^2} - {a^2}} } \right)e}}$
- D
$B = \frac{{2m{v_0}}}{{\left( {\sqrt {{b^2} - {a^2}} } \right)e}}$
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- [AIEEE 2005]
If the magnetic field is parallel to the positive $y-$axis and the charged particle is moving along the positive $x-$axis (Figure), which way would the Lorentz force be for
$(a)$ an electron (negative charge),
$(b)$ a proton (positive charge).
If the magnetic field is parallel to the positive $y-$axis and the charged particle is moving along the positive $x-$axis (Figure), which way would the Lorentz force be for
$(a)$ an electron (negative charge),
$(b)$ a proton (positive charge).