Answer:
[tex]E=6.01\times 10^{-3}\ J[/tex]
Explanation:
It is given that,
Side of a square loop, l = 24 cm = 0.24 m
Resistance of loop, R = 6.1 ohms
Initial magnetic field is 0.665 T and final magnetic field is 0 as the field is removed in 40 ms
We need to find the electrical energy dissipated in this process.
Due to change in magnetic field, the loop will induce a voltage. The induced voltage is given by :
[tex]V=-\dfrac{dB}{dt}\\\\V=\dfrac{BA}{t}[/tex]
If I is induced current then,
[tex]V=IR[/tex]
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{BA}{tR}[/tex]
Power is given by voltage times current. So,
[tex]P=\dfrac{(BA)^2}{(t^2R)}[/tex]
Now, energy is given by the product of power and time. So,
[tex]E=\dfrac{(BA)^2}{(t^2R)}\times t\\\\E=\dfrac{(BA)^2}{(tR)}[/tex]
Now putting all the values in above formula. So,
[tex]E=\dfrac{(0.665\times (0.24)^2)^2}{(40\times 10^{-3}\times 6.1)}\\\\E=6.01\times 10^{-3}\ J[/tex]
So, the electrical energy of [tex]6.01\times 10^{-3}\ J[/tex] is dissipated in this process.
The electrical energy dissipated throughout this process will be "6.01 × 10⁻³ J".
According to the question,
Square loop's side, l = 24 cm or,
= 0.24 m
Loop's resistance, R = 6.1 ohms
Initial magnetic field = 0.665 T
Final magnetic field = 0
We know the relation,
→ V = - [tex]\frac{dB}{dt}[/tex]
= [tex]\frac{BA}{t}[/tex]
Also we know,
Current, V = IR
I = [tex]\frac{V}{R}[/tex]
= [tex]\frac{BA}{tR}[/tex]
Now, Energy, E = [tex]\frac{(BA)^2}{t^2R}[/tex] × t or,
= [tex]\frac{(BA)^2}{tR}[/tex]
By substituting the values,
= [tex]\frac{(0.665\times (0.24)^2)^2}{40\times 10^{-3}\times 6.1}[/tex]
= 6.01 × 10⁻³ J
Thus the response above is correct.
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