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Home » 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?
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?
CBSE | Class 12 | Electromagnetic Induction | NCERT | Physics
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?

Answer –       

Maximum emf induced is given = 0.603 V

Average emf induced is given by= 0 V

Maximum current in the coil = 0.0603 A

Power loss (average) is given = 0.018 W

(It is the power coming from external rotor)

Circular coil radius is given by r = 8 cm = 0.08 m

Area of the coil is then given by –

A=\pi {{r}^{2}}=\pi \times {{(0.08)}^{2}}{{m}^{2}}

Number of turns on the coil is given by N = 20

Angular speed is

\omega =50rad/\sec

Strength of magnetic field is given by

B=3\times {{10}^{-2}}T

Total resistance produced by the loop is given by R = 10Ω

Maximum emf induced is given by the expression –

e=N\omega AB=20\times 50\times \pi \times {{(0.08)}^{2}}\times 3\times {{10}^{-2}}

e=0.603V

The maximum emf induced in the coil is 0.603 V.

The average emf induced in the coil is zero over a full cycle.

Maximum current is given by the expression –

I=\frac{e}{R}=\frac{0.603}{10}

    \[I=0.0603A\]

Average power because of the Joule heating is given by the expression –

    \[P=\frac{el}{2}=\frac{0.603\times 0.0603}{2}\]

    \[P=0.018W\]

The torque generated by the current induced in the coil opposes the coil’s natural rotation. To maintain the coil rotating continually, we need a source of torque that opposes the emf’s torque, so the rotor acts as an external agent. As a result, the external rotor is the source of dissipated power.

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