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bekas [8.4K]
3 years ago
13

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 μ

T. 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?
Physics
1 answer:
blagie [28]3 years ago
7 0

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.

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