"Some pacemakers employ magnetic reed switches to enable doctors to change their mode of operation without surgery. A typical reed switch can be switched from one position to another with a magnetic field of 5.0×10−4T. Part A What current must a wire carry if it is to produce a 5.0×10−4T field at a distance of 0.50 m?"

The current the wire must carry is 1250 A

Explanation:

Given;

strength of magnetic field, B = 5.0 × 10⁻⁴ T

distance from the wire, r = 0.50 m

The strength of magnetic field on the current carrying wire is given as;

where;

B is the magnetic field strength

μ₀ is permeability of free space, = 4π x 10⁻⁷ T.m/A

I is the current on the wire

r is the distance from the wire

Make current "I" the subject of the formula;

Therefore, the current the wire must carry is 1250 A

answer: its c lololo

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg