An electron is moving in the vicinity of a long, straight...

An electron is moving in the vicinity of a long, straight wire that lies along the $x$ -axis. The wire has a constant current of 9.00 A in the $-x$ -direction. At an instant when the electron is at point $(0,0.200 \mathrm{~m}, 0)$ and the electron's velocity is $\overrightarrow{\boldsymbol{v}}=\left(5.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\right) \hat{\imath}-\left(3.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath},$ what is the force that the wire exerts on the electron? Express the force in terms of unit vectors, and calculate its magnitude.

Key Concepts

Magnetic Field of a Current-Carrying Conductor

This concept involves determining the magnetic field generated by a steady current flowing through a conductor. Using Ampère's law, the magnetic field at a distance r from an infinitely long, straight wire is given by B = (?0I)/(2?r), where ?0 is the permeability of free space. The direction of the field is determined by the right?hand rule, which relates the direction of current flow to the circular magnetic field lines surrounding the conductor.

Lorentz Force Law

The Lorentz force law describes the force experienced by a moving charged particle in a magnetic field. It is given by F = q(v × B), where q is the charge, v is the velocity vector, and B is the magnetic field vector. This law shows that the force is perpendicular to both the velocity of the charge and the magnetic field, and its magnitude depends on the magnitude of the charge, the speed of the particle, and the sine of the angle between the velocity and the field.

Vector Cross Product and Right-Hand Rule

The vector cross product is a mathematical operation used to determine both the magnitude and direction of the force in the Lorentz force law. The direction of the resulting vector is determined by the right-hand rule, which involves orienting the fingers of the right hand in the direction of the first vector (velocity) and curling them toward the second vector (magnetic field). The thumb then points in the direction of the resultant force, ensuring correct vector orientation.

Transcript

0:00 High.

00:02 In the given problem here there is a straight current carrying conductor of infinite length which has been kept along the x -axis like this.

00:20 This current this conductor is carrying a current along negative x -axis.

00:29 The magnitude of this current is given as 9 .00 ampere.

00:35 Then there is a point point an observation point let it be p along positive y axis here this is y axis here this is x axis so the x xx is will be perpendicular to the plane of paper somewhere like this now the distance of this observation point has been given as 0 .200 meter so using biotent severed slot the magnetic field at this point will be having a magnitude given by mu not by 4 pi into 2 i by r so plugging in all known values for mu not by 4 pi this is 10 dash per minus 7 tesla meter per ampere multiplied by 2 times of current which is 9 .00 ampere divided by distance which is 0 .200 meter so canceling this meter and ampere this 0 .2 will make it 20 then canceling this this is 10 so finally the magnitude of this magnetic field at the observation point comes out to be 9 into 10 dash bar minus 6 tesla now using right -hand screw rule right -hand screw rule direction of this magnetic field will be into the plane of paper or we can say this magnetic field will be along the negative xx.

02:22 So, the direction of b is along negative x -axis.

02:31 So in vector form, this magnetic field may also be written as minus 9 .0 into 10 dash power minus 6, tesla, k, k, k.

02:44 Now the electron, an electron was moving at that instant at this point p, which was having a velocity given by 5 .00 into 10 days per 4 meter per second, i cap minus 3 .00 into 10 is per 4 meter per second j cap.

03:15 So the magnetic lawrence force experienced by this electron will be given by expression q into v cross v and for q this is a charge over an electron which is minus e into v cross p so if we take for the time thing we we take this electronic charge as e only then in place of this v cross b for v this is 5 or we can say this is 5 i cap minus 3 j cap into 10 tish par 4 meter per second cross multiplied by b which was minus 9 into 10 dash per minus 6 tesla into k cap so now expanding this bracket and taking all other common things out we get this minus e then 10 dash to power 4 meter per second and then this minus 9 into 10 dash power minus 6 tesla that can also be taken as a common out so finally we will be having 5 i cap cross k cap minus 3 j cap cross k cap minus 3 j cap cap now using the rules for vector product for the cross product i cross k will be equal to minus j and j cross k will be equal to plus i and now finally we can use the value of this electronic charge also here so this force magnetic laureates force here will come out to be minus 1 .6 into 10 dash par minus 19 kulam multiplied by 9 and this minus minus into minus will become plus so this is plus and here this is into 10 dash 4 into 9 into 10 dash for minus 6 so this is into 10 dash power 4 into 10 to the power minus 9 and into 10 is to power minus 6 so sorry this is 10 dash 4 into 9.

05:43 Here this is 9 and then into 10 dash the bar minus 6...

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