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

Answer:

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;

$B = \frac{\mu_o I}{2\pi r}$

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;

$B = \frac{\mu_o I}{2\pi r} \\\\I = \frac{2\pi rB}{\mu_o} \\\\I = \frac{2\pi *0.5*5*10^{-4}}{4\pi *10^{-7} }\\\\I = 1.25 *10^3 \ A\\\\I = 1250 \ A$

Therefore, the current the wire must carry is 1250 A