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 5.0x10-4 T. What current must a wire carry if it is to produce a 5.0x10-4 T field at 0.50m?I = ? kA

Answer 1

1.25 kA

Step-by-step explanation:

The magnitude of the magnetic field produced by a current-carrying wire is given by:

`B=(\mu_0 I)/(2 \pi r)`

where

`\mu_0 = 1.26 \cdot 10^(-6) H/m` is the vacuum permeability

`I` is the current in the wire

r is the distance from the wire at which the field is calculated

In this problem, we have:

`B = 5.0\cdot 10^(-4) T` is the intensity of the magnetic field

`r = 0.50 m` is the distance from the wire

Re-arranging the equation, we can find the intensity of the current in the wire:

`I=(2 \pi r B)/(\mu_0)=(2 \pi (0.50 m)(5.0\cdot 10^(-4)T))/(1.26\cdot 10^(-6) H/m)=1246 A=1.25 kA`