Subjective Type

A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of $$ 200 V m^{-1}$$ makes the path straight . Find the charge / mass ratio of the particle.

Solution

Let charge on particle is $$q$$ and mass of particle is $$m$$. Moving with velocity $$v$$, perpendicular to magnetic field $$(B)$$ now. Radius of circle related as $$\cfrac { mV^{ 2 } }{ r } =qVB$$
$$ V=\cfrac { q }{ m } Br$$
If electrostatics force balance electromagnetic force however $$q\vec { E } =qVB$$
$$\Rightarrow \vec { E } =VB=\cfrac { q }{ m } Br(B)$$
$$\Rightarrow \vec { E } =(\cfrac { q }{ m } )B^{ 2 }r$$
$$\cfrac { q }{ m } =\cfrac { \vec { E } }{ B^{ 2 }r } =\cfrac { 200 }{ (0.4)^{ 2 }(\cfrac { 1\times 10^{ -2 } }{ 2 } ) } =\cfrac { 400 }{ 0.16\times { 10 }^{ -2 } } $$
$$ \cfrac { q }{ m } =250\times { 10 }^{ 3 }=250k$$

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