Question

Problem 6. Two infinitely long wires are arranged side by side. The current I, is pointing out of the page. The current I1 is pointing into the page. The distance between the wires is d. 

Let the magnitude of the currents be equal. I = I) Find the direction of the total magnetic field at the following coordinates. Give the answer as a positive or negative unit vector i, j, or k.

23. At (2d, 0) 

(A) rightward (B) up (C) out (D) leftward (E) down 

24. At (1/2d, 1/2d) 

(A) rightward (B) up (C) out (D) leftward (E) down 

Now let 12 = 31, for the following questions. 

25. What is the direction of the magnetic force vector F_2,3 acting on currently due to the current I₁? 

(A) Up (B) Down (C) Left (D) Right (E) Out 

26. Find the magnitude of the force per length (in units of 10^-4 N/m) acting on I₂ due to I₁ if I₁ = 3.0 A and d=2.0 cm. 

(A) 3.5 (B) 8.5 (C) 3.8 (D) 2.7 (E) 5.9 

27. Find the location "x" (in cm) where the net magnetic field is zero. Again let d 2.0 cm. 

(A)-1.0 (B) -0.5 (C) 0.5 (D) 1.0 (E) 1.5 

Consider the magnetic field due to both I1 and I2 at a field point that is a distance 2d above i1. Remember to let the magnitude of I2 be three times the magnitude of I1 for all the questions below. (I2 = 31) 

28. Find the magnitude of the x-component of the magnetic field at the coordinate (0, 2d ). Give a simplified expression in terms of I and d. 

29. An electron at the coordinate (0, 2d) moves with a speed 3.0 m/s in the positive y-direction. Find the magnitude of the acceleration that the electron would experience at the field point. Let I = 3.0 A and d=2.0 cm. Give an answer in units of 10⁶ m/s². 

(A) 1.0 (B) 4.7 (C) 2.7 (D) 3.4 (E) 5.8 

30. What would be the direction of the force that would act on the electron? 

(A) Up (B) Down (C) Left (D) In (E) Out

Answer 1

Solution:

(23). The magnetic field at (2d,0) is given by

Thus, the direction of magnetic field at (2d,0) is " down"

(24). Similarly, the magnetic field at (d/2, d/2) is given by

Hence, the direction of magnetic field at (d/2, d/2) is " up"

(25). Magnetic field on I2 due to I1 is given by

Magnetic force on I2 due to I1 is given by

So, the direction of magnetic force on I2 due to I1 is " Right".

(26). Magnetic force per unit length on I2 due to I1 is given by

Given thea,

Now substituting these values in above equation, we get

Obtion (D) is correct.