Respuesta :
Explanation:
Given that,
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1∘ from the vertical, [tex]\theta=15.1^{\circ}[/tex]
The magnitude of the magnetic field B changes in time according to the equation :
[tex]B(t)=3.75+2.75 t-7.05 t^2[/tex]
Radius of the loop, r = 0.21 m
We need to find the magnitude of the induced emf in the loop when t=5.63 s. The induced emf is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA\cos \theta)}{dt}[/tex]
B is magnetic field
A is area of cross section
[tex]\epsilon=A\dfrac{-dB}{dt}\\\\\epsilon=\pi r^2\dfrac{-d(3.75+2.75 t-7.05 t^2)}{dt}\times \cos\theta\\\\\epsilon=\pi r^2\times(2.75-14.1t)\times \cos\theta[/tex]
At t = 5.63 seconds,
[tex]\epsilon=-\pi (0.21)^2\times(2.75-14.1(5.63))\times \cos(15.1)\\\\\epsilon=10.25V[/tex]
So, the magnitude of induced emf in the loop when t=5.63 s is 10.25 V.
The EMF generated at time t = 5.63 is 10.18V.
Magnetic flux and EMF:
Given a horizontal circular wire loop with a radius, r = 0.21m.
A time-dependent magnetic field B(t) = 3.75 + 2.75t -7.05t².
At an angle of θ = 15.1° to the area of the loop.
The magnetic flux passing through the loop is given by:
Ф = B(t)Acosθ
where A = πr² is the are of the loop.
Since the magnetic field is time-dependent, the magnetic flux through the loop changes with time, therefore an EMF is generated in the loop, given by:
[tex]E=-\frac{d\phi}{dt}\\\\E =-\frac{dB(t)}{dt}Acos\theta\\\\E=-\pi r^2cos\theta\frac{d}{dt}[ 3.75 + 2.75t -7.05t^2] \\\\E=\pi r^2cos\theta[14.10t-2.75]\\\\[/tex]
At time t = 5.63s
[tex]E=3.14\times(021)^2\times cos15.1\times[14.1\times5.63-2.75][/tex]
E = 10.18V
Learn more about magnetic flux:
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