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saveliy_v [14]
3 years ago
10

A coil 3.85 cm radius, containing 450 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1

0−2 T/s )t+( 2.60×10−5 T/s4 )t4. The coil is connected to a 700-Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.
Find the magnitude of the induced emf in the coil as a function of time.
Physics
2 answers:
Nimfa-mama [501]3 years ago
6 0

Answer:

0.025V + (0.000218V/s³) t³

Explanation:

Parameters given:

Radius of coil, r = 3.85 cm = 0.0385 m

Number of turns, N = 450

Magnetic field, B = ( 1.20×10^(−2) T/s )t + (2.60×10^(−5) T/s4 )t^4.

The magnitude of Induced EMF is given as:

E = N * A * dB/dt

Where A is the area of the coil

First, we differentiate the magnetic field with respect to time:

dB/dt = 0.012 + 0.000104t³

Therefore, EMF will be:

E = 450 * 3.142 * (0.012 + 0.000104t³)

E = 2.096(0.012 + 0.000104t³)

E = 0.025V + (0.000218V/s³)t³

rjkz [21]3 years ago
4 0

Answer:

The magnitude of the induced emf is E=0.02511V+(0.000217\frac{V}{s^{3} } )t^{3}

Explanation:

The area is:

A=\pi r^{2}

r = 3.85 cm = 0.0385 m

A=\pi *0.0385^{2} =0.00465m^{2}

The induce emf is:

E=NA\frac{dB}{dt}

Replacing

E=450*0.00465*((\frac{d(1.2x10^{-2})t }{dt} )+(\frac{d(2.6x10^{-5})t^{4}  }{dt} )\\E=2.0925*(1.2x10^{-2} +0.000104t^{3} )\\E=0.02511V+(0.000217\frac{V}{s^{3} } )t^{3}

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