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)
S safeer

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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