Let’s look at a simple application of Faraday\'s law. In (Figure 1) , the magnet
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Let’s look at a simple application of Faraday's law. In (Figure 1) , the magnetic field in the region between the poles of the electromagnet is uniform at any time but is increasing at the rate of 0.020T/s. The area of the conducting loop in the field is 120 cm2 , and the total circuit resistance, including the meter, is 5.0 ? . Find the magnitudes of the induced emf and the induced current in the circuit.
A)Suppose we change the apparatus so that the magnetic field increases at a rate of 0.13 T/s , the area of the conducting loop in the field is 0.035 m2 , and the total circuit resistance is 6.5 ? . Find the magnitude of the induced emf.
B)Find the magnitude of the induced current in the circuit.
C Chegg.com × MasteringPhysics: HMWK #7 × Po field%253 - https://session.masteringphysics.com/myct/itemView?assignmentProblemID-60804522&offset-next; Chegg Spring 2016 Signed in as Miranda AasenHelp Close HMWK #7 Item 2 Resources previous 2 of 25 next» Part A Practice Problem: Let's look at a simple application of Faraday's law. In Al Figure 1), the magnetic field in the region between the Suppose we change the apparatus so that the magnetic field increases at a rate of 0.13 T/s, the area of the conducting loop in the field is 0.035 m2 , and the total circuit resistance is 6.5 Find the magnitude of the induced emf Express your answer in volts to two significant figures. on poles of the electromagnet is uniform at any time but is increasing at the rate of 0.020T/s. The area of the conducting loop in the field is 120 cm2 , and the total le Te circuit resistance, including the meter, is 5.02. Find the magnitudes of the induced emf and the induced current in the circuit Figure 1 1 of 1 xXb A = 120 cm2 = 0.012 m2 Total resistance in circuit and meter = 5.0 Submit My Answers Give Up Incorrect Try Again, no points deducted Part B-Practice Problem: Top companies Global citiesExplanation / Answer
given that
dB/dt = 0.020 T/s
A = 120 *10^(-4) m^2
R= 5 ohm
magnetic flux is:
/PHI = A*B
The induced emf is:
E = (deltaPHI)/ delta t
E = (A* deltaB)/delta t
E= A*dB/dt
E = (120 *10^(-4) )*( 0.020) = 0.24 cV
The induced current is:
I = E / R = 0.0024 V /5 ohm = 0.48 mA
(a)
magnetic flux is:
/PHI = A*B
The induced emf is:
E = (deltaPHI)/ delta t
E = (A* deltaB)/delta t
E= A*dB/dt
E = (0.035 )*( 0.13) = 4.5 mV
(b)
The induced current is:
I = E / R
I = 0.0045 / 6.5 = 0.7 mA
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