Physics, asked by mochikim343, 1 day ago

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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Answered by aadhi202007
0

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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

This is the answer

This is the answer

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