Question

A small flat search coil of area $5 cm^{2}$ with $140$ closely wound turns is placed between the poles of a powerful magnet producing magnetic field $ 0.09 T$ and then quickly removed out of the field region. Calculate
(a) magnetic flux through the coil, and
(b) emf induced in the coil.

Answer 1

Hint: When a coil is placed in the magnets, magnetic field produced. Due to magnetic field, magnetic flux generated and induced current generated in the coil so, emf also generated in the coil of opposite polarity. Emf generated is equal to the rate of change of magnetic flux with respect to the time.

Complete answer:
Given area of coil, A=$5 cm^{2} = 5 \times 10^{-4} m^{2}$
Number of turns, N = $140$
Magnetic field, B = $ 0.09 T$
a) Magnetic flux, $\phi = NBA$
Putting values of N, B, A in the above formula of magnetic flux, we get
 $\phi =140 \times 0.09 \times 5 \times 10^{-4}$
$\phi = 6.3 \times 10^{-3}$Wb
Hence, magnetic flux through the coil is $6.3 \times 10^{-3}$ Wb.
b) Induced emf in the coil-
$e =- \dfrac{d\phi}{dt}$
 As the coil is removed quickly within a millisecond.
$e = -\dfrac{\Delta \phi}{\Delta t}$
Here, the magnetic field is removed, flux will be zero.
$\Delta \phi = (0 - 6.3 \times 10^{-3}) Wb$
$\Delta t = 1 \times 10^{-3}$s
$e = \dfrac{ 6.3 \times 10^{-3}}{10^{-3}} = 6.3 V$

Hence, emf induced in the coil is $6.3 V$.

Note: The magnetic field can generate a current. The magnetic field change provided by the coil produces an emf and a current. Emf generated in the coil is of opposite direction and equal to the rate of change of flux. Due to the opposite direction, a negative sign comes into play.