Answer:
Radius of the loop is 0.18 m or 18 cm
Explanation:
Given :
Current flowing through the wire, I = 45 A
Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T
Number of turns in circular wire, N = 1
Consider R be the radius of the circular wire.
The magnetic field at the center of the current carrying circular wire is determine by the relation:
Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.
Substitute the suitable values in the above equation.
R = 0.18 m
(a) -0.211 m
At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is
The projection of the length of the pendulum along the vertical direction is
the full length of the pendulum when the mass is at the lowest position is
L = 36.1 cm
So the y-displacement of the mass is
(b) 0.347 J
The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to
where we have
m = 168 g = 0.168 kg is the mass of the pendulum
g = 9.8 m/s^2 is the acceleration due to gravity
is the vertical displacement of the pendulum
So, the work done by gravity is
And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.
(c) Zero
The work done by a force is:
where
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and the displacement
In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula
and so the work done is zero.