Question

. A circular wire loop 40 cm in diameter has 100 Ohm resistance and lies in a horizontal plane. A uniform magnetic field points vertically downward, and in 25 ms it increases from 5 mT to 55 mT. Find the magnetic flux through the loop at (a) the beginning and (b) the end of 25 ms period. (c) What is the loop current during this time

Answer 1

Answer:

(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A

Explanation:

Given that,

The diameter of the loop, d = 40 cm

Radius, r = 20 cm

Initial magnetic field, B = 5 mT

Final magnetic field, B' = 55 mT

Initial magnetic flux,

$\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb$

Final magnetic flux,

$\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb$

Due to change in magnetic field an emf will be generated in the loop. It is given by :

$\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V$

Let I be the current in the loop. We can find it using Ohm's law such that,

$\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A$

Hence, this is the required solution.