Home
Class 12
PHYSICS
A fan blade of length 2a rotates with fr...

A fan blade of length 2a rotates with frequency f cycles per second perpendicular to a magnetic field B. Find the p.d between the centre and the end of the blade.

Text Solution

AI Generated Solution

The correct Answer is:
To find the potential difference (p.d.) between the center and the end of a rotating fan blade in a magnetic field, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Given Information:** - Length of the fan blade = 2a - Frequency of rotation = f (cycles per second) - Magnetic field = B (perpendicular to the plane of rotation) 2. **Identify the Area Swept by the Blade:** - The fan blade rotates around its center, so the distance from the center to the end of the blade is 'a'. - The area swept by the blade during one complete rotation (one cycle) is given by the formula for the area of a circle: \[ \text{Area} = \pi r^2 = \pi a^2 \] 3. **Calculate the Time Period (T):** - The time period (T) is the reciprocal of frequency (f): \[ T = \frac{1}{f} \] 4. **Use Faraday's Law of Electromagnetic Induction:** - According to Faraday's law, the electromotive force (emf, ε) induced in a circuit is equal to the rate of change of magnetic flux through the circuit: \[ \epsilon = -\frac{d\Phi}{dt} \] - The magnetic flux (Φ) through the area swept by the blade is: \[ \Phi = B \cdot \text{Area} = B \cdot \pi a^2 \] 5. **Calculate the Rate of Change of Magnetic Flux:** - The rate of change of magnetic flux as the blade rotates is: \[ \frac{d\Phi}{dt} = B \cdot \frac{d(\pi a^2)}{dt} \] - Since the area is constant, we can express the change in time as: \[ \frac{d\Phi}{dt} = B \cdot \pi a^2 \cdot \frac{1}{T} = B \cdot \pi a^2 \cdot f \] 6. **Calculate the Induced EMF (p.d.):** - Substituting this into Faraday's law gives: \[ \epsilon = -B \cdot \pi a^2 \cdot f \] - The negative sign indicates the direction of the induced emf according to Lenz's law, but for the magnitude of the potential difference (p.d.), we can ignore the negative sign: \[ \text{p.d.} = B \cdot \pi a^2 \cdot f \] ### Final Answer: The potential difference (p.d.) between the center and the end of the blade is: \[ \text{p.d.} = B \cdot \pi a^2 \cdot f \]
Promotional Banner

Topper's Solved these Questions

  • ELECTRICAL INSTRUMENTS

    SL ARORA|Exercise Problem for self practice|81 Videos
  • ELECTROMAGNETIC WAVES

    SL ARORA|Exercise Problem for self-practice|21 Videos

Similar Questions

Explore conceptually related problems

A fan blade of length 2 a rotates with frequency f cycle' per second perpendicular to magnetic field B . Then, potenial difference between centre and end of blade is

A fan blade of length 1//sqrt(pi) metre rotates with frequncy 5 cycle per second perpendicular to a magnetic field 10 tesla. What is potential diffrence between the centre and the end of blade ?

A copper rod of length L rotates with an angular with an angular speed 'omega' in a uniform magnetic field B. find the emf developed between the two ends of the rod. The field in perpendicular to the motion of the rod.

A metallic metre sitck moves with a velocity of 2 ms ^(-1) in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.2 T. Find the emf indcued between the ends of the stick.

A conducting rod of unit length moves with a velocity of 5m/s in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.4T . Find the emf induced between the ends of the stick.

A metal rod of length 2 m is rotating with an angular velocity of 100 rad//sec in a plane perpendicular to a uniform magnetic field of 0.3 T . The potential difference between the ends of the rod is

SL ARORA-ELECTROMAGNETIC INDUCTION-All Questions
  1. A wire 40 cm long bent into a rectangular loop 15 cmxx5 cm is placed ...

    Text Solution

    |

  2. The magnetic flux through a coil perpendicular to its plane and direct...

    Text Solution

    |

  3. A fan blade of length 2a rotates with frequency f cycles per second pe...

    Text Solution

    |

  4. The magnetic flux linked with a coil of N turns and resistance R chang...

    Text Solution

    |

  5. A metal disc of radius 200 cm is rotated at a constant angular speed o...

    Text Solution

    |

  6. A gramophone disc of brass of diameter 30 cm rotates horizontally at t...

    Text Solution

    |

  7. A long solenoid with 15 turns per cm has small loop of area 2.0 cm^(2)...

    Text Solution

    |

  8. An air-cored solenoid of length 50cm and area of cross-sectional 28 cm...

    Text Solution

    |

  9. A closed coil consists of 500 turns has area 4 cm^2 and a resistance o...

    Text Solution

    |

  10. A jet plane is moving at a speed of 1000 km h^(-1). What is the potent...

    Text Solution

    |

  11. A straight conductor of length 0.4 m is moved with a speed of 7 m/s pe...

    Text Solution

    |

  12. A horizontal telephone wire 1 km long is lying along east-west in eart...

    Text Solution

    |

  13. A horizontal wire 24 cm long falls in the field of flux density 0.8 T...

    Text Solution

    |

  14. A straight rod 2m long is placed in an aeroplane in the east-west dire...

    Text Solution

    |

  15. Shows a square loop having 100 turns, an area of 2.5 xx10^(-3) m^2 an...

    Text Solution

    |

  16. A rod closing the circuit shown in Fig moves along a U shaped wire at ...

    Text Solution

    |

  17. Fig. 12.32 shows a long, rectangular, conducting loop of with l, mass ...

    Text Solution

    |

  18. A coil has 50 turns and its area is 500 cm^(2). It is rotating at the ...

    Text Solution

    |

  19. Calculate the maximum emf induced in a coil of 100 turns and 0.01 m^(2...

    Text Solution

    |

  20. A magnetic flux of 8xx10^(-4) weber is linked with each turn of a 200 ...

    Text Solution

    |