Answer:
a. 5.09 × 10⁻⁷ T b. 2.67 × 10⁻⁶ T
Explanation:
a. What is the horizontal component of the Earth's magnetic field?
The magnetic field due to the circular coil, B = μ₀Ni/L where μ₀ = 4π × 10⁻⁷ H/m, N = number of turns of coil = 7, i = current = 0.540 A and L = length of coil = πND where N = number of turns of coil = 7 and D = diameter of coil = 30.0 cm = 0.30 m.
So, B = μ₀Ni/L
= μ₀Ni/πND
= μ₀i/πD
Substituting the values of the variables into the equation, we have
B = μ₀i/πD
= 4π × 10⁻⁷ H/m × 0.540 A/(π × 0.30 m)
= 4 × 10⁻⁷ H/m × 0.540 A/0.30 m
= 2.16 × 10⁻⁷ HA/m/0.30 m
= 7.2 × 10⁻⁷ T
Since this magnetic field is at 45° to the horizontal, its horizontal comoonent equals the horizontal component B' of the earth's magnetic field.
So, Bcos45° = B'
B' = (7.2 × 10⁻⁷ T)cos45°
B' = 5.09 × 10⁻⁷ T
b. What is the total magnitude of the Earth's magnetic field at this location?
Since the angle of dip at this location is 11.0° and B" is the magnetic field at this location, the horizontal component of B" = B'
So B"sin11.0° = 5.09 × 10⁻⁷ T
B" = 5.09 × 10⁻⁷ T/sin11.0°
B" = 5.09 × 10⁻⁷ T/0.1908
B" = 26.68 × 10⁻⁷ T
B" = 2.668 × 10⁻⁶ T
B" ≅ 2.67 × 10⁻⁶ T
