Question

A 105‑turn circular coil of radius 2.41 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is connected to a 14.7 Ω resistor to create a closed circuit. During a time interval of 0.167 s, the magnetic field strength decreases uniformly from 0.481 T to zero. Find the energy, in millijoules, that is dissipated in the resistor during this time interval.

Answer 1

Given : N = 105 ; r = 2.41 cm (or 0.0241 m) ; R = 14.7 ; t=0.167 s

B = 0.481 T

Solution:

Change in magnetic flux can be given as :

= B×A

Where , A = r²

i.e. A = (0.0241m)² = 18.24×10^-4 m²

Hence = (0.481 T)(18.24×10^-4 m²) = 8.77×10^-4 Tm²

The emf can be calculated as :

= 0.551 V

The current is given by :

= 0.0375 A

The energy dissipated in the resistor is given by :

E= I²Rt

= (0.0375 A)²(14.7 )(0.167 s)

= 0.00345 J or say 3.45 mJ

Answer : 3.45 mJ