RaySix

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at November 2nd 2023, 5:10:53 pm.

Question:

A long straight wire carries a current of 2.5 A in the positive x-direction. The wire is positioned in the xy-plane and the current flows through it in the positive x-direction. A positively charged particle with a charge of +3.5 μC and a mass of 6.4 × 10^-23 kg is moving in the xy-plane with a velocity of 3.0 × 10^5 m/s in the positive y-direction.

a) Calculate the magnetic field at a point P, located at coordinates (2.0 m, 0.0 m, 0.0 m), due to the wire.

b) What is the magnitude and direction of the magnetic force experienced by the positively charged particle at point P?

c) If the magnetic field at point P was found to be zero, what could be the orientation of the wire with respect to the x-axis?

Answer:

a) To calculate the magnetic field at point P due to the wire, we can use the formula for the magnetic field produced by a current-carrying wire:

$B = rac{u_0}{4i} rac{I}{r}$

where:

B is the magnetic field
$I$ is the current in the wire
$r$ is the distance from the wire to the point where we want to calculate the magnetic field
$mu_0$ is the permeability of free space and has a value of $4i imes 10^{-7} T dot m/A$

Given that the current in the wire is 2.5 A and the distance from the wire to point P is 2.0 m, we can substitute these values into the formula to find the magnetic field at point P:

$B = rac{(4i imes 10^{-7} T dot m/A) dot (2.5 A)}{2.0 m}$

Simplifying, we find:

$B = 3.14 imes 10^{-6} T$

Therefore, the magnetic field at point P due to the wire is 3.14 × 10^-6 T directed out of the xy-plane.

b) To find the magnitude and direction of the magnetic force experienced by the positively charged particle at point P, we can use the formula for the magnetic force on a charged particle moving in a magnetic field:

$ec{F} = qec{v} imes ec{B}$

where:

$q$ is the charge of the particle
$v$ is the velocity vector of the particle
$B$ is the magnetic field vector

Given the charge of the particle as +3.5 μC (3.5 × 10^-6 C), the velocity of the particle as 3.0 × 10^5 m/s in the positive y-direction, and the magnetic field at point P as 3.14 × 10^-6 T directed out of the xy-plane, we can substitute these values into the formula to find the magnetic force:

$ec{F} = (3.5 imes 10^{-6} C) dot (3.0 imes 10^5 m/s) dot (3.14 imes 10^{-6} T)$

Simplifying, we find:

$ec{F} = 3.30 imes 10^{-6} N$

The magnitude of the magnetic force experienced by the particle at point P is 3.30 × 10^-6 N. Since the velocity of the particle is in the positive y-direction and the magnetic field is out of the xy-plane, the direction of the magnetic force can be determined using the right-hand rule. Curl the fingers of your right hand from the positive y-direction to the positive x-direction and your thumb points in the direction of the magnetic force. Thus, the direction of the magnetic force is in the positive z-direction.

c) If the magnetic field at point P was found to be zero, it means that the wire is not oriented in a way to produce any magnetic field at point P. In this case, the wire must be either parallel or antiparallel to the x-axis.

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Magnetic Field due to Current-Carrying Wire