A potential difference of $$600$$ V is applied across the plates of a parallel plate condenser. The separation between the plates is $$3$$ mm. An electron projected vertically, parallel to the plates, with a velocity of $$2$$×$$106$$ $$ms-1$$ moves undeflected between the plates. Find the magnitude of the magnetic field in the region between the condenser plates. (Neglect$$ the edge effects. Charge of the $$electron=$$ $$-1.6$$×$$10$$−$$19C$$ $$)

Question

A potential difference of $$600$$ V is applied across the plates of a parallel plate condenser. The separation between the plates is $$3$$ mm. An electron projected vertically, parallel to the plates, with a velocity of $$2$$×$$106$$ $$ms-1$$ moves undeflected between the plates. Find the magnitude of the magnetic field in the region between the condenser plates. (Neglect$$ the edge effects. Charge of the $$electron=$$ $$-1.6$$×$$10$$−$$19C$$ $$)

Infinity Learn · Accepted Answer

The force on electron will be toward the left plane due to electric field and will be equal to  $$Fe=eE$$.For no deflection, electric force should be balanced by magnetic force.$$eE=evB$$⇒$$B=Ev=V/dvWhere$$ V $$=potential$$ difference between the plates, $$andd=distance$$ between the $$platesB=600/3$$×$$10$$−$$32$$×$$106=6003$$×$$10$$−$$3$$×$$2$$×$$106$$ ⇒$$B=0.1$$ T .

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