Question

A loop antenna of area A = 3.04 cm^2 and resistance R = 6.66 μΩ is perpendicular to a uniform magnetic field of magnitude 18.4 μT. The field magnitude drops to zero in 5.43 ms. How much thermal energy is produced in joules in the loop by the change in field?

Answer 1

Answer:

Thermal energy, $H=8.57\times 10^{-10}\ J$

Explanation:

It is given that,

Area of loop antenna, $A=3.04\ cm^2=0.000304\ m^2$

Resistance, $R=6.66\ \mu \Omega=6.66\times 10^{-6}\ \Omega$

Magnetic field, $B=18.4\ \mu T=18.4\times 10^{-6}\ T$

The field magnitude drops to zero in 5.43 ms, $t=5.43\times 10^{-3}\ s$

Due to change in magnetic field, an EMF will induced in the loop which is given by :

$\epsilon=\dfrac{d(BA)}{dt}$

Also, $\epsilon=IR$

$IR=\dfrac{d(BA)}{dt}$

$I=\dfrac{d(BA)}{Rdt}$

So, $I=\dfrac{18.4\times 10^{-6}\times 0.000304}{6.66\times 10^{-6}\times 5.43\times 10^{-3}}$

I = 0.154 A

Thermal energy produced is given by,

$H=I^2Rt$

$H=(0.154)^2\times 6.66\times 10^{-6}\times 5.43\times 10^{-3}$

$H=8.57\times 10^{-10}\ J$

So, the thermal energy produced by changing field is $8.57\times 10^{-10}\ J$. Hence, this is the required solution.