Answer:
a
The force acting on one of the sides the square loop is 169.6 N
b
The torque acting on the loop is ZERO
Explanation:
From the question we are given that
permeability of free space [tex]\mu_0[/tex] [tex]= 4\pi*10^{-7} T\cdot m/A[/tex]
Number of turns of the solenoid per unit length , n = 30 turns per cm
= 3000 turns per m
The current flowing through the solenoid [tex]I_1[/tex] = 15A
Length of the side of the loop, L = 2 cm
= 2 cm [[tex]\frac{1*10^{-2}m}{1cm}[/tex]]
= [tex]2*10^{-2} m[/tex]
Angle , [tex]\theta =90[/tex]°
The current flowing through the wire [tex]I_2 = 0.15A[/tex]
Generally The formula for Magnetic field inside of a solenoid is
[tex]B = \mu_0nI_1[/tex]
Where
[tex]\mu_0[/tex] is the permeability of free space
n is the number of turns of the solenoid per unit length
[tex]I_1[/tex] is the current through the solenoid
Also the formula for force acting on one of the sides of a square loop is
[tex]F = BI_2Lsin \theta[/tex]
Where [tex]I_2[/tex] is the current through the wire ,
L is the length of the sides of the loop,and
[tex]\theta[/tex] is the angle between B and [tex]I_2[/tex]
Substituting the given values into the first formula
[tex]B = (4 \pi *10^{-7})(3000)(15)[/tex]
[tex]= 5.65*10^{-2} T[/tex]
To obtain the force acting on the loop by substituting the appropriate value into the second formula
[tex]F = (5.65*10^{-2}(0.15)(2*10^{-2}(sin90)))[/tex]
[tex]169.6 \mu N[/tex]
The torque acting on the lop is zero because the
The forces on the loop are acting in the plane of the loop
These force are perpendicular to the sides
and their direction is away from the inner side of the loop
the only effect the orientation of these forces can cause is stretching of the loop
This way by which the forces are oriented cannot cause any rotation
