Question

"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?"

Answer 1

The current the wire must carry is 1250 A

Step-by-step 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;

`B = (\mu_o I)/(2\pi r)`

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;

`B = (\mu_o I)/(2\pi r) \\\\I = (2\pi rB)/(\mu_o) \\\\I = (2\pi *0.5*5*10^(-4))/(4\pi *10^(-7) )\\\\I = 1.25 *10^3 \ A\\\\I = 1250 \ A`

Therefore, the current the wire must carry is 1250 A