For MRI, a patient is slowly pushed in a time of 10 ~s within the coils of the magnet where magnetic field is B=2.0 ~T . If the patient's trunk is 0.8 ~m in circumference, the induced emf around the patient's trunk is

Correct answer: C. 10.18 × 10^ -3 ~V Change in magnetic field in 10 ~s=2.0 ~T As =(-d φ)/(d t)=-A (d B)/(d t) ( φ=B A) Circumference of patient's trunk, 2 π r=0.8 ~m (given) r=(0.8)/(2 π) m=(0.4)/(π) m Areaofcrosssection, A=π r^2=π((0.4)/(π))^2=(0.16)/(π) m^2 | |=(0.16)/(π) × (2)/(10) ~V 10.18 × 10^ -3 ~V

AIIMS 2017

Exam: AIIMS
Year: 2017
Subject: Physics
Chapter: Electromagnetic Induction

AIIMS 2017 · Physics · Electromagnetic Induction

For MRI, a patient is slowly pushed in a time of 10 ~s within the...

For MRI, a patient is slowly pushed in a time of

10 \mathrm{~s}

within the coils of the magnet where magnetic field is

B=2.0 \mathrm{~T}

. If the patient's trunk is

0.8 \mathrm{~m}

in circumference, the induced emf around the patient's trunk is

Options

A $10.18 \times 10^{-2} \mathrm{~V}$
B $9.66 \times 10^2 \mathrm{~V}$
C $10.18 \times 10^{-3} \mathrm{~V}$
D $1.51 \times 10^{-2} \mathrm{~V}$

Correct answer

C. $10.18 \times 10^{-3} \mathrm{~V}$

Step-by-step solution

Change in magnetic field in

10 \mathrm{~s}=2.0 \mathrm{~T}

\varepsilon=\frac{-d \phi}{d t}=-A \frac{d B}{d t}

(\because \phi=B A)

Circumference of patient's trunk,

2 \pi r=0.8 \mathrm{~m}

(given)

\therefore \quad r=\frac{0.8}{2 \pi} \mathrm{m}=\frac{0.4}{\pi} \mathrm{m}

Areaofcrosssection,

A=\pi r^2=\pi\left(\frac{0.4}{\pi}\right)^2=\frac{0.16}{\pi} \mathrm{m}^2

\therefore|\varepsilon|=\frac{0.16}{\pi} \times \frac{2}{10} \mathrm{~V} \approx 10.18 \times 10^{-3} \mathrm{~V}

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