Answer:
Explanation:
Given that;
horizontal circle at a rate of 2.33 revolutions per second
the magnetic field of the Earth is 0.500 gauss
the baton is 60.1 cm in length.
the magnetic field is oriented at 14.42°
we wil get the area due to rotation of radius of baton is
The formula for the induced emf is
B is the magnetic field strength
substitute
The magnetic field of the earth is oriented at 14.42
we plug in the values in the equation above
so, the induce EMF will be
Answer:
(A) 0.279N at angle 38.02°
(B) 0.701N
(C) 14.19°
Explanation:
(A) The net force on q3 is given as:
F = Fxi + Fyj
Fx is the x component of the force
Fy is the y component of the force
Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0
Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx
First let us find y and angle x from the diagram.
Using Pythagoras theorem,
y² = 0.3² + 0.4²
y² = 0.25
y = 0.5m
Using SOHCAHTOA to find x,
sinx = 0.4/0.5
x = 53.13°
Electrostatic force, F is given as:
F = kqQ/r²
Where k = Coulumbs constant
F(1,3) = (k*q1*q3) / r²
F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)
F(1, 3) = 0.288N
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = 0.45N
Therefore,
Fx = -0.288cos36.87 + 0.45
Fx = 0.22N
Fy = 0.288cos53.13
Fy = 0.172N
=> F = 0.22i + 0.172j
The magnitude of the force will be
F(mag) = √(0.22² + 0.172²)
F(mag) = 0.279N
The direction of the force makes will be
tanθ = Fy/Fx
tanθ = 0.172/0.22 = 0.781
θ = 38.02° to the x axis.
(B) q2 = - 2.0 * 10^(-6)
This implies that:
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = -0.45N
Therefore,
Fx = -0.288cos36.87 - 0.45
Fx = -0.68N
Fy = 0.172N
=> F = - 0.68i + 0.172j
The magnitude of the force will be
F(mag) = √((-0.68)² + 0.172²)
F(mag) = 0.701N
(C) The direction of the force makes will be
tanθ = 0.172/0.68
θ = 14.19° to the x axis
