Respuesta :
Answer:
[tex]-1.14 \cdot 10^{-3} V[/tex]
Explanation:
The induced emf in the loop is given by Faraday's Newmann Lenz law:
[tex]\epsilon = - \frac{d \Phi}{dt}[/tex] (1)
where
[tex]d\Phi[/tex] is the variation of magnetic flux
[tex]dt[/tex] is the variation of time
The magnetic flux through the coil is given by
[tex]\Phi = NBA cos \theta[/tex] (2)
where
N = 6 is the number of loops
A is the area of each loop
B is the magnetic field strength
[tex]\theta =15^{\circ}[/tex] is the angle between the direction of the magnetic field and the normal to the area of the coil
Since the radius of each loop is r = 5.00 cm = 0.05 m, the area is
[tex]A=\pi r^2 = \pi (0.05 m)^2=0.0079 m^2[/tex]
Substituting (2) into (1), we find
[tex]\epsilon = - \frac{d (NBA cos \theta)}{dt}= -(NAcos \theta) \frac{dB}{dt}[/tex]
where
[tex]\frac{dB}{dt}=0.0250 T/s[/tex] is the rate of variation of the magnetic field
Substituting numbers into the last formula, we find
[tex]\epsilon = -(6)(0.0079 m^2)(cos 15^{\circ})(0.0250 T/s)=-1.14 \cdot 10^{-3} V[/tex]
Answer:
Induced emf, [tex]\epsilon=-1.13\times 10^{-3}\ volts[/tex]
Explanation:
It is given that,
Number of circular loop, N = 6
A uniform magnetic field perpendicular to its surface is held stationary at an angle of 15 degrees to the ground.
Radius of the loop, r = 5 cm = 0.05 m
Change in magnetic field, [tex]\dfrac{dB}{dt}=0.025\ T/s[/tex]
Due to the change in magnetic field, an emf will be induced. Let E is the induced emf in the coil. it is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
[tex]\phi[/tex] = magnetic flux
[tex]\epsilon=\dfrac{d(NBA\ cos\theta)}{dt}[/tex]
[tex]\epsilon=-NA\dfrac{d(B)}{dt}[/tex]
[tex]\epsilon=6\times \pi (0.05)^2\times 0.025\times cos(15)[/tex]
[tex]\epsilon=-1.13\times 10^{-3}\ volts[/tex]
So, the induced emf in the loop is [tex]1.13\times 10^{-3}\ volts[/tex] . Hence, this is the required solution.
