A long straight wire AB carries a current 4A.A proton travels at 4×10^6m/s parallel to the wire,0.2 from it and in a direction opposite to the current .Find the force which the magnetic field of current exerts on the proton
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,r=0.2mI=4A
The current carrying wire AB ProducesamagneticfieldatthepositionofpointP,B=μoI2πr ,whichisperpendiculartoplaneofpaperdownward.Theprotonmovesperpendiculartothismagneticfield,Hence,experiencesthemagneticforce,Fm=qvBMagneticfield,B=μoI2πrB=4π×10−7×42π×0.2=4×10−6TMagneticforce,Fm=qvBv=4×106m/sFm=1.6×10−19×4×106×4×10−6Fm=25.6×10−19NAccordingtoflemingslefthandrule,thedirectionofforceisawayfromwireAB.
The current carrying wire AB ProducesamagneticfieldatthepositionofpointP,B=μoI2πr ,whichisperpendiculartoplaneofpaperdownward.Theprotonmovesperpendiculartothismagneticfield,Hence,experiencesthemagneticforce,Fm=qvBMagneticfield,B=μoI2πrB=4π×10−7×42π×0.2=4×10−6TMagneticforce,Fm=qvBv=4×106m/sFm=1.6×10−19×4×106×4×10−6Fm=25.6×10−19NAccordingtoflemingslefthandrule,thedirectionofforceisawayfromwireAB.
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The current exerted on the proton is 25.6×10−19NA
Radius of the wire = r = 0.2m (Given)
Current in the wire I = 4A (Given)
The wire carrying the current produces magnetic field at = P
Thus,
,B=μoI/2πr, ( perpendicular to the plane of paper downwards)
Hence, Fm=qvB
Magnetic field, B = μoI/2πrB
= 4π×10−7×4/2π×0.2
= 4×10−6T
Magnetic force = Fm = qvB
v = 4 × 10`6m/s
Fm = 1.6 × 10−19×4×106×4×10−6
Fm=25.6 × 10−19NA
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