Question

A rectangular loop of sides $10cm$ and $5cm$ with a cut is stationary between the pole pieces of an electromagnet. The magnetic field of the magnet is normal to the loop. The current feeding the electromagnet is reduced so that the field decreased from its initial value of $0.3T$ at the rate of $0.02T/s$. If the cut is joined and the loop has a resistance of $2.0\Omega$, the power dissipated by the loop as heat is

• A. $5nW$
• B. $4nW$
• C. $3nW$
• D. $2nW$

Answer 1

The correct option is A

$5nW$

Step 1: Given data:

Area of the rectangular loop,$A=10x5=50c{m}^2=50x{10}^{-4}{m}^2$

Change in magnetic intensity,$dB/dt=0.02T/s$

Resistance of the coil, $R=2\Omega$

Let the current be $i$ and Power Dissipation be $P$.

Step 2: Finding the Power Dissipated $P$ by the loop as heat:

We know that,

Electromagnetic Force $e=$current $\times$Total Load Resistance

$e=iR\Rightarrow i=\frac{e}{R}$

Electromagnetic Force in terms of magnetic field is,

$e=A\frac{dB}{dt}$ (where, $A$ is Area and $B$ is Magnetic Field.)

Using, $i=\frac{e}{R}$

$$\begin{gathered} i=\frac{A{\displaystyle \frac{dB}{dt}}}{R} \\ \Rightarrow i=\frac{50x{10}^{-4}x0.02}{2} \\ \Rightarrow i=5x{10}^{-5}A \end{gathered}$$

We know that,

Power, $\begin{array}{rcl}P& =& i^{2}R\end{array}$

$\begin{array}{rcl}& \Rightarrow & P={\left(5x{10}^{-5}\right)}^{2}x2\\ & \Rightarrow & P=25x{10}^{-10}x2\\ & \Rightarrow & P=50x{10}^{-10}\\ & \Rightarrow & P=5nW\end{array}$

The power dissipated by the loop as heat is $\begin{array}{rcl}& & 5nW\end{array}$.

Hence, option A is correct.