Question

The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic field used to align the spins, with proportionality constant 5 hz/T. If the strength of the applied field is known to be 20 T plus or minus 3 T, which of the following correctly describes the uncertainty in the INVERSE frequency (1/frequency)? Choose 1 answer: a. 3/2000s b. 3/5s c. 1/15sd. 4s

Answer 1

Answer:

The inverse frequency is $\dfrac{3}{80}\ s$

Explanation:

Given that,

Magnetic field = 20 T

Proportionality constant = 5 Hz/T

Change in magnetic field = 3 T

We know that,

$B=\dfrac{K}{\dfrac{1}{\omega}}$

We need to calculate the inverse frequency

Using formula of frequency

$\Delta(\dfrac{1}{\omega})=\dfrac{\Delta B}{k\times(\dfrac{1}{\omega^2})}$

$\Delta(\dfrac{1}{\omega})=\dfrac{k\times\Delta B}{B^2}$

Put the value into the formula

$\Delta(\dfrac{1}{\omega})=\dfrac{3\times5}{(20)^2}$

$\Delta(\dfrac{1}{\omega})=\dfrac{3}{80}\ s$

Hence, The inverse frequency is $\dfrac{3}{80}\ s$