Question

PLEASE ANSWER HELP!!! PLEASE ANSWER BOTH PARTS

1) A charged particle moves with velocity 1.5 km/s making angle 50° with the external magnetic field of magnitude of 0.6 T (see figure below). The field exerts the 3.5 N force on the charge. The direction of the force is out of the page/screen. What is the magnitude and sign of the charge?

The magnetic force would be zero if:
Х 50° В

1. v was parallel to the field;
2. v was directed out of page;
3. v was directed into the page
4. v=0;
5. A and D;
6. B and C

2)

1. A wire of 0 m length carries the current I₁ = 25 A. Another wire of the same length, parallel to the first one carries the current I₂ = 15 A. As a result, the wires attract each other with the force F = 5 mN. What is the distance between the wires? Are the currents of the same or opposite directions?

What of the following statements is/are correct?

1. The current in one wire create electric field, which excerts a force on the second wire with the current;
2. The current in one wire create magnetic field, which excerts a force on the second wire with the current;
3. The force between the wires is inverse-proportional to the distance between wires;
4. The force between the wires is inverse-proportional to the squared distance between wires;
5. A and D;
6. B and C.

Answer 1

1) From the right hand thumb rule you can see that the cross product vector of (v x B) is into the page/screen. To make it out of the screen the charge has to be of negative sign.

Now the magnitude of the force can be given as

|F| = |qvBsin50|

=> 3.5N = q x 1500m/s x 0.6T x sin50

=> q = 5.0766 mC

Since the magnetic force is given as F = q( v x B)

thus if v is parallel to B (v || B ) then v x B = 0 as angle between them is 0 or 180.

which makes the magnetic force F as 0.

Now if v = 0 then you can clearly see force F = 0

Thus, option E) is correct.

2) I think you missed the length of wire though to mention assume it to be L, Now the magnetic force applied by any wire will be

B = $\mu_o$ I/2$\pi$d

Now assuming the first is producing the magnetic field such that it is pulling the second wire towards itself so, we write that force on it will be

F = I₂LB₁

=> F = $\mu_o$ I₁I₂L/2$\pi$d

equating this to the force given we have;

=> 5 x 10^-3 N = 2 x 10^-7 x 25A x 15A x L/d

=> d = 0.015L

Now i don't think L will be 0m that makes no sense of this question, so please check it and use it in the above equation to obtain the value of d which will be the answer for the distance.

Since the wires are attracting each other by right hand thumb rule you can see that both wires have to carry current in same direction in order to attract each other.

Well, the current in one wire creates a magnetic field which exerts a forceon the second wire with the current and that is inversely proportional to the distance between them as can be seen from the equation above.

So option F will be correct.