Question

The magnitude of the magnetic field in a magnetic resonance imaging (MRI) machine can be as great as B = 2.0 T . Under normal circumstances, this field cannot be shut off by just flipping a switch. Instead the magnitude needs to be carefully decreased to zero. In an emergency, however, the magnet can be "quenched" so that B reduces to zero in 20 s. Such a quench can cost thousands of dollars and likely damages the magnets. Assume that the magnetic field exists inside a cylinder of radius R = 400 mm and length ℓ = 300 mm .

1. How much magnetic potential energy is dissipated when the magnetic field is quenched in this way?2. What is the average rate at which energy is dissipated?

Answer 1

Answer

given,

B = 2 T

time = 20 s

Radius = R = 400 mm

L= 300 mm

apply the formula for magnetic energy density

$\eta_0 = \dfrac{energy}{volume}$

$\eta_0 = \dfrac{B^2}{2\mu}$

$\eta_0 = \dfrac{2^2}{2\times 4 \pi 10^{-7}}$

energy density = 1.59 x 10⁶ J/m³

Now , in the cylinder  ,

Energy   = energy density  x volume

E = 1.59 x 10⁶  x π x r² x h

E = 1.59 x 10⁶  x π x 0.4² x 0.5

E = 4 x 10⁵  J

now, average rate of energy dissipated

rate of dissipation of energy = Energy/time  

                                                = 4 x 10⁵ /20

                                                = 2 x 10⁴ W

the average rate at which energy is dissipated is 2 x 10⁴ W

Answer 2

1. The energy be "4 × 10⁵ J".

2. The average rate be "2 × 10⁴ W"

Magnetic field

According to the question,

Time, t = 20 s

Radius, R = 400 mm

Length, L = 300 mm

We know the formula,

Magnetic energy density,

→ $\eta_0$ = $\frac{Energy}{Volume}$

      = $\frac{B^2}{2 \mu}$

By substituting the values, we get

      = $\frac{(2)^2}{2\times 4 \pi 10^{-7}}$

      = 1.59 × 10⁶ J/m³

Now,

1.

Energy will be:

= Energy density × Volume

= 1.59 × 10⁶ × π × (0.4)² × 0.5

= 4 × 10⁵ J

2.

We know,

The rate of dissipation = $\frac{Energy}{Time}$

                                       = $\frac{4\times 10^5}{20}$

                                       = 2 × 10⁴ W

Thus the above response is correct.

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