Answer :
Answer:
B = 0.204T
Explanation:
To find the value of the magnetic force you use the following formula:
[tex]F_B=ILBsin\theta[/tex]
I: current of the copper rod = 45A
B: magnitude of the magnetic field
L: 0.78m
you assume that magnetic field B and current I are perpendicular between them.
The magnetic force must be, at least, equal to the friction force, that is:
[tex]F_{f}=F_{B}\\\\\mu N=\mu Mg=ILB\\\\B=\frac{\mu Mg}{IL}[/tex]
M: mass of the rod = 1.20kg
μ: coefficient of static friction = 0.61
g: gravitational acceleration constant = 9.8m/s^2
By replacing the values of the parameters you obtain:
[tex]B=\frac{(0.61)(1.20kg)(9.8m/s^2)}{(45A)(0.78m)}=0.204T[/tex]
Answer:
The magnitude of the smallest magnetic field is [tex]B = 0.1744 \ T[/tex]
Explanation:
From the question we are told that
The mass of the copper is [tex]m = 1.20 kg[/tex]
The distance of separation for the rails is [tex]d = 0.78 \ m[/tex]
The current is [tex]I = 45 A[/tex]
The coefficient of static friction is [tex]\mu = 0.61[/tex]
The force acting along the vertical axis is mathematically represented as
[tex]F = mg - F_y[/tex]
Where [tex]F_y[/tex] is the force acting on copper rod due to the magnetic field generated this is mathematically represented as
[tex]F_y = I * d * B_1[/tex]
The magnetic field here is acting towards the west because according to right hand rule magnetic field acting toward the west generate a force acting in the vertical axis
So the equation becomes
[tex]F = mg - I * d * B_1[/tex]
Here [tex]B_1 = B sin \theta[/tex]
[tex]F = mg - I * d * Bsin(\theta )[/tex]
The in the horizontal axis is mathematically represented as
[tex]F_H = ma + F_x[/tex]
Since the rod is about to move it acceleration is zero
Now [tex]F_x[/tex] is the force acting in the horizontal direction due to the magnetic field acting downward this is because a according to right hand rule magnetic field acting downward generate a force acting in the horizontal positive horizontal direction. this mathematically represented as
[tex]F_H = 0 + I * d * B_2[/tex]
So the equation becomes
[tex]F_H = I * d * B_2[/tex]
Here [tex]B_2 = B cos \theta[/tex]
[tex]F_H = I * d * Bcos (\theta)[/tex]
Now the frictional force acting on this rod is mathematically represented as
[tex]F_F = \mu * F[/tex]
[tex]F_F = \mu * (mg -( I * d * Bsin(\theta )))[/tex]
Now when the rod is at the verge of movement
[tex]F_H = F_F[/tex]
So [tex]I * d * Bcos (\theta) = \mu * (mg -( I * d * Bsin(\theta )))[/tex]
=> [tex]B = \frac{\mu mg }{I * d (cos \theta + \mu (sin \theta ))}[/tex]
Now [tex]\theta[/tex] is the is the angle of the magnetic field makes with the vertical and the horizontal and this can be mathematically evaluated as
[tex]\theta = tan^{-1} (\mu )[/tex]
Substituting value
[tex]\theta = tan^{-1} ( 0.61 )[/tex]
[tex]\theta = 31.38^o[/tex]
Substituting values into the equation for B
[tex]B = \frac{0.61 (1.20) (9.8)}{(45) (0.78) (cos (31.38) + 0.61 (sin (31.38)) )}[/tex]
[tex]B = 0.1744 \ T[/tex]