6.6 A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10^–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

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Payal Gupta | Contributor-Level 10

4 months ago

6.6 Given,

Radius of the circular coil, r = 8.0 cm = 0.08 m

Area of the coil, A = $π r^{2}$ = $π \times {0.08}^{2}$ = 20.106 $\times 10^{- 3}$ $m^{2}$

Number of turns in the coil, n = 20

Angular speed, $ω$ = 50 rad/s

Magnetic field strength, B = 3.0 $\times 10^{- 2}$ T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as e = n $ω A B$ = 20 $\times 50 \times$ 20.106 $\times 10^{- 3} \times$ 3.0 $\times 10^{- 2}$ = 0.603 V

Over a full cycle, the average emf induced in the coil is zero.

Maximum current is given by, $I_{m a x}$ = $\frac{e}{R}$ = $\frac{0.603}{10}$ = 0.0603 A

Average power loss due to Joule heating is given by, $P_{l o s s}$ = $\frac{e I_{m a x}}{2}$ = 0.018 W

The current induced in the coil p

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6.6 Given,

Radius of the circular coil, r = 8.0 cm = 0.08 m

Area of the coil, A = $π r^{2}$ = $π \times {0.08}^{2}$ = 20.106 $\times 10^{- 3}$ $m^{2}$

Number of turns in the coil, n = 20

Angular speed, $ω$ = 50 rad/s

Magnetic field strength, B = 3.0 $\times 10^{- 2}$ T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as e = n $ω A B$ = 20 $\times 50 \times$ 20.106 $\times 10^{- 3} \times$ 3.0 $\times 10^{- 2}$ = 0.603 V

Over a full cycle, the average emf induced in the coil is zero.

Maximum current is given by, $I_{m a x}$ = $\frac{e}{R}$ = $\frac{0.603}{10}$ = 0.0603 A

Average power loss due to Joule heating is given by, $P_{l o s s}$ = $\frac{e I_{m a x}}{2}$ = 0.018 W

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent; it must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

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6.6 Given,Radius of the circular coil, r = 8.0 cm = 0.08 mArea of the coil, A = <math><mi>π</mi><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = <math><mi>π</mi><mo>×</mo><mi></mi><msup><mrow><mrow><mn>0.08</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 20.106 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math>Number of turns in the coil, n = 20Angular speed, <math><mi>ω</mi></math> = 50 rad/sMagnetic field strength, B = 3.0 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> TResistance of the loop, R = 10 ΩMaximum induced emf is given as e = n <math><mi></mi><mi>ω</mi><mi></mi><mi>A</mi><mi>B</mi></math> = 20 <math><mo>×</mo><mn>50</mn><mo>×</mo></math> 20.106 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup><mo>×</mo></math> 3.0 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> = 0.603 VOver a full cycle, the average emf induced in the coil is zero.Maximum current is given by, <math><msub><mrow><mrow><mi>I</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></math> = <math><mfrac><mrow><mrow><mi>e</mi></mrow></mrow><mrow><mrow><mi>R</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>0.603</mn></mrow></mrow><mrow><mrow><mn>10</mn></mrow></mrow></mfrac></math> = 0.0603 AAverage power loss due to Joule heating is given by, <math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mi>l</mi><mi>o</mi><mi>s</mi><mi>s</mi></mrow></mrow></msub></math> = <math><mfrac><mrow><mrow><mi>e</mi><msub><mrow><mrow><mi>I</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math> = 0.018 WThe current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent; it must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

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