A circular coil that has 110 turns and a radius of 14 cm is in a magnetic field

Question

A circular coil that has 110 turns and a radius of 14 cm is in a magnetic field that has a magnitude of 0.0650 T directed perpendicular to the coil. What is the magnetic flux through the coil? A circular coil has N = 110 turns and a radius of 0.14 m. There is an external magnetic field B = 0.0650 T directed perpendicularly through the coil. The magnetic flux through the coil is equal to Phi_0 = NBA cos(Theta), where A is the cross-sectional area of the coil and Theta is the angle between the magnetic field and the area vector of the coil. The magnetic field at the coil is increased steadily to 0.100 T over a time interval of 0.5 s. What is the magnitude of the emf induced in the coil during the time interval? The external magnetic field is steadily increased from 0.0650 T to 0.100 T over Delta t = 0.5 s; the induced potential in the coil as a result of this change is |epsilon| = |-Delta Phi_B/Delta t|.V

Explanation / Answer

(a)
N = 110 turns,
Magnetic flux through the coil is given by,
m = NBAcos()
As the Magnetic Field is perpdicular to the coil, Angle between A * B is 0o.
m = NBA * cos(0)
m = NBA
m = 110 * 0.0650 * 3.14*0.14^2
m = 0.44 Wb

(b)

Magitude of the emf induced in the coil,
e = - N* d/dt
e = - 110 */t

t = 0.5 s
= 110 * 3.14*0.14^2 (0.1 - 0.0650) = 0.2369 wb

e = - 110 *(0.2369/0.5)
e = - 52.18 V

Magnitude of the emf induced, e = 52.18 V

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