Ncert Exercises & Solutions

Q 11. Let us assume that the loop in question number 4 is stationary or constant but the current source which is feeding the electromagnet which is producing the magnetic field is slowly decreased. It was having an initial value of 0.3 T and the rate of reducing the field is 0.02 T / sec. If the cut is joined to form the loop having a resistance of $1.6 Ω$ . Calculate how much power is lost in the form of heat? What is the source of this power?

The sides of the rectangular loop are 8 cm and 2 cm.
Hence, the area of the rectangular wire loop,
$\begin{array}{l} A = length \times width \\ = 8 \times 2 = 16 c m^{2} \\ = 16 \times 10^{- 4} m^{2} \end{array}$
The initial value of the magnetic field, B' = 0.3T
Rate of decrease of the magnetic field, $\frac{d B}{d t} = 0.02 T / s$
Emfdeveloped in the loop is given as: $e = \frac{d \emptyset}{d t}$
Where, dnull Change in flux through the loop area = AB
$\begin{array}{l} ∴ e = \frac{d (A B)}{d t} = \frac{A d B}{d t} \\ = 16 \times 10^{- 4} \times 0.02 = 0.32 \times 10^{- 4} V \end{array}$
The resistance of the loop, R= 1.6 $Ω$
The current induced in the loop is given as:
$\begin{array}{l} i = \frac{e}{R} \\ = \frac{0.32 \times 10^{- 4}}{1.6} = 2 \times 10^{- 5} A \end{array}$
The power dissipated in the loop in the form of heat is given as:
$\begin{array}{l} P = i^{2} R \\ = (2 \times 10^{- 5})^{2} \times 1.6 \\ = 6.4 \times 10^{- 10} W \end{array}$
The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

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