Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;

Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
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Answer:
See below
Explanation:
Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N
no work is done by this force
Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N
work of friction = 7.55 * 2 m = 15.1 j
Force Downplane = mg sin 53 = 62.61 N
work = 62.61 * 2 = 125.22 j
Net Force downplane = force downplane - force friction = 55.06 N
net Work = force * distance = 55.06 N * 2 M = 110.12 j