A square current loop 4.9 cm on each side carries a 520 mA current. The loop is in a 1.0 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30∘ away from the field direction.

What is the magnitude of the torque on the current loop? 2 sig fig.

Torque = 0.25Nm

Explanation:

To calculate the magnitude of torque, first we must calculate the lever arm. The lever arm is perpendicular distance from the axis of rotation to the line of the action of the force.

Lever arm = r sin theta

Where r = 4.9cm= 0.049m

Theta = 30°

Lever arm = 0.049 sin30° = 0.245m

Magnitude of torque = rfsintheta

f = 1.0T

Magnitude of torque= 0.245×1.0 = 0.245Nm

To 2 significant figure=0.25Nm

3

step-by-step explanation:

10 - 2.8 = 7.2

2 2/5 = 2.4

7.2 / 2.4 = 3

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.