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

Thermal energy,
`H=8.57* 10^(-10)\ J`

Step-by-step 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* 10^(-6)\ \Omega`

Magnetic field,
`B=18.4\ \mu T=18.4* 10^(-6)\ T`

The field magnitude drops to zero in 5.43 ms,
`t=5.43* 10^(-3)\ s`

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

`\epsilon=(d(BA))/(dt)`

Also,
`\epsilon=IR`

`IR=(d(BA))/(dt)`

`I=(d(BA))/(Rdt)`

So,
`I=(18.4* 10^(-6)* 0.000304)/(6.66* 10^(-6)* 5.43* 10^(-3))`

I = 0.154 A

Thermal energy produced is given by,

`H=I^2Rt`

`H=(0.154)^2* 6.66* 10^(-6)* 5.43* 10^(-3)`

`H=8.57* 10^(-10)\ J`

So, the thermal energy produced by changing field is
`8.57* 10^(-10)\ J`. Hence, this is the required solution.