Question

Two long, straight wires, one above the other, are seperated by a distance d = 1.53 cm and are parallel to the x−axisx−axis. Let the +y−axis+y−axis be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current 34 A in the +x−direction+x−direction. Find the magnetic force on a negative point charge 79 μCμC moving with velocity 3.49×105 m/s in the +y-direction, when the charge is:

Answer 1

Answer:

a)   $F_x = F_y = F_x = 0 \ \ N$

b)  $F_x= - 0.05718 \ N ; F_y = 0 \ N ; F_z= 0 \ N$

Explanation:

Given that:

q = -79 μC

q = - 79×10⁻⁶ C

d = 1.53 cm

d = 0.0153 m

I = 34 A

v = 3.49×10⁵ m/s

$\bar{V} = v \hat{j}$

The force on charge q is given by

$\bar{ F } = q (\bar {v} * \bar{B})$

a) At midway (A) , the B will be :

$\bar{B} =\bar{B_1} + \bar{B_2}$

$B_1$  and  $B_2$ will be equally the same in respect to their magnitude but opposite direction at point A.

So; $\bar{|B_1|} = |B_2| = \frac{\mu_oI}{2 \pi d/2}$

where $|B| =0$

∴  $\bar {|F|} = 0$

∴ $F_x = F_y = F_x = 0 \ \ N$

b)

At a distance d/2 cm above the upper wire:

$\bar{B} =\bar{B_1} + \bar{B_2}$

where:

$B_1 =B_2 = \frac{\mu_o I }{2 \pi ( d + d/2 )}$   ; upward to the plane of paper

∴ $\frac{\mu_o I}{2 \pi} \ \ [\frac{2}{3/d} +\frac{2}{d}] \ \ \hat{k}$

B = $\frac{\mu_o I}{ \pi \ d} \ \ [\frac{1}{3} +1] \ \ \hat{k}$

B = $\frac{\mu_o I}{ \pi \ d} \ [\frac{4}{3} ] \ \hat{k}$

B = $\frac{7 \pi * 10^{-7} *34}{\pi*0.0153}*\frac{4}{3} \ \hat{k}$

B = 2.074 × 10⁻³T  $\hat {k}$

∴ $\bar {F} = q ( \bar{v} + \bar {B})$

$\bar {F} = q ( \bar{v} + B)j*k\\\\\bar {F} = q ( \bar{v} + B) \bar {i}$

$F = -79*10^{-6}*3.49*10^5*2.074*10^{-3}$

$F = - 0.05718 \ N \ \hat {i}$

$F_x= - 0.05718 \ N ; F_y = 0 \ N ; F_z= 0 \ N$