An MRI technician moves his hand from a region of very low magnetic field streng

Question

An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.10 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.21 cm, and it takes 0.370 s to move it into the field. (a) What average current is induced in the ring if its resistance is 0.0100 ? (Enter the magnitude in amperes.) A (b) What average power is dissipated (in W)? W (c) What average magnetic field is induced at the center of the ring? (Enter the magnitude in teslas.) T (d) What is the direction of this induced magnetic field relative to the MRI's field? parallel antiparallel The magnitude is zero. † An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.10 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.21 cm, and it takes 0.370 s to move it into the field. (a) What average current is induced in the ring if its resistance is 0.0100 ? (Enter the magnitude in amperes.) A (b) What average power is dissipated (in W)? W (c) What average magnetic field is induced at the center of the ring? (Enter the magnitude in teslas.) T (d) What is the direction of this induced magnetic field relative to the MRI's field? parallel antiparallel The magnitude is zero. † An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.10 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.21 cm, and it takes 0.370 s to move it into the field. (a) What average current is induced in the ring if its resistance is 0.0100 ? (Enter the magnitude in amperes.) A (b) What average power is dissipated (in W)? W (c) What average magnetic field is induced at the center of the ring? (Enter the magnitude in teslas.) T (d) What is the direction of this induced magnetic field relative to the MRI's field? parallel antiparallel The magnitude is zero. †

Explanation / Answer

a) Magnitude of induced emf, Vin = /t = (B/t)(d2/4)

Induced current, Iin = Vin/R = d2B / 4Rt

=> Iin = * (2.21 * 10-2)2 * (2.10 - 0) / (4 * 0.0100 * 0.370) = 0.218 A

b) Average power dissipated = Iin2R = 0.2182 * 0.0100 = 4.75 * 10-4 W

c) Induced magnetic field at the center of the ring,

B = oIin/d = (4 * 10-7) * 0.218 / (2.21 * 10-2) = 1.24 * 10-5 T

d) Since the magnetic flux through the ring increases, the magnetic field due to the induced current will oppose the external magnetic field i.e. antiparallel to the 2.10 T field.

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