A long, straight wire carrying a 5 A current is strapped on your arm which is us

Question

A long, straight wire carrying a 5 A current is strapped on your arm which is used to study your blood flow. At one instant, a hydrogen ion (proton) in your vein, 4.0 mm from the wire, travels at 1.5 times 10^3 m/s parallel to the wire and in the opposite direction to the current (See Fig. 2 below) Find the: (a) Magnitude and direction of the total magnetic field. B. that is acting at point P due to the magnetic field produced by the wire and the proton, which has a force 4.8 times 10^-20 N acting on it. (b) Magnitude and direction of the force acting on the hydrogen ion due to the current passing through the wire.

Explanation / Answer

a)
at point P

B_wire = mue*I/(2*pi*d)

= 4*pi*10^-7*5/(2*pi*6*10^-3)

= 1.67*10^-4 T (into the page)

B_Ion = (mue/(4*pi))*(q*v/r^2)

= (4*pi*10^-7/(4*pi))*(1.6*10^-19*1.5*10^3/0.002^2)

= 6*10^-18 T (out of the page)

Bnet = B_wire - B_IOn

= 1.67*10^-4 T (since B_Ion is very small) <<<----Answer

b) magnetic field at the location of proton,

B = mue*I/(2*pi*d)

= 4*pi*10^-7*5/(2*pi*4*10^-3)

= 2.5*10^-4 T (into the page)

magnetic force acting on the proton, F = q*v*B*sin(90)

= 1.6*10^-19*1.5*10^3*2.5*10^-4

= 6*10^-20 N

direction : Towards right (away from the wire)

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