Write the relation for the force f acting on a charge carrier q moving with a velocity v
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Solution :
F→=q(V→×B→)F→=q(V→×B→)
(i) F→F→ is perpendicular to both V-V-and B→B→ .
If ds-ds- is the instantaneous displacement of the change ds→ds→ is also perpendicular to F→F→
Work done by this force = increase in kinetic energy
w=F→.ds→w=F→.ds→
=Fscos90=Fscos90
But cos90=0cos90=0
w=0w=0
Hence increase in kinetic energy is zero.
(ii) power = F→.v→F→.v→
=Fvcos90=Fvcos90
=0=0
Power =0
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F→=q(V→×B→)F→=q(V→×B→)
(i) F→F→ is perpendicular to both V-V-and B→B→ .
If ds-ds- is the instantaneous displacement of the change ds→ds→ is also perpendicular to F→F→
Work done by this force = increase in kinetic energy
w=F→.ds→w=F→.ds→
=Fscos90=Fscos90
But cos90=0cos90=0
w=0w=0
Hence increase in kinetic energy is zero.
(ii) power = F→.v→F→.v→
=Fvcos90=Fvcos90
=0=0
Power =0
If like please thank❤❤❤
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