A long straight wire XY carries a current of 2A. A proton travels at 5 x 10 m/s parallel to the wire as shown in the diagram, 0.2 m away from it. The force that the magnetic field.
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A long straight wire AB carries a current of 4 A. A proton P travels at parallel to the wire 0·2 m from it and in
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A long straight wire AB carries a current of 4 A. A proton P travels at 4\times 10^{6}ms^{-1} parallel to the wire 0·2 m from it and in a direction opposite to the current as shown in the figure. Calculate the force which the magnetic field due to the current carrying wire exerts on the proton. Also specify its direction.
Answers (1)
Ssafeer
The magnetic field at point P due to
Current carrying straight wire AB,
B=\frac{\mu _{o}}{2\pi }\frac{I}{r}\; \; \; \; \; \; \; (i)
Where 'I' is current
'r' is the distance at point P from the wire.
Now,
Force acting on moving proton in the magnetic field;
F=Bqv \; \sin \theta --------(ii)
Where, 'B' is magnetic field
'q' is charge
'v' is velocity
from equation (i) putting the value of B in equation (ii), we get
F=\frac{\mu _{o}}{2\pi }\frac{I}{r}q\; v\; \sin \theta
We have given that;
v = 4\times 10^{6}m/s
q = 1.6\times 10^{-19}C
I = 4A
r = 0.2m
Therefore:-
F=\frac{2\times 10^{-7}\times 4\times 1.6\times 10^{-19}\times 4\times 10^{6}\times \sin 90^{o}}{0.2}\; \; \; \; \; \; \; \;
F=\frac{32\times 1.6\times 10^{-20}}{0.2}
F=2.56\times 10^{-18}N
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