Home
Class 12
PHYSICS
A simple pendulum with bob of mass m and...

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle `2theta`. The earth's magnetic field component in the direction perpendicular to swing is B. The maximum potential difference induced across the pendulum is

A

a.`2BLsin((theta)/(2))sqrt(gL)`

B

b.`BL sin((theta)/(2))sqrt((gl))`

C

c..`BL sin ((theta)/(2)).(gL)^(3//2)`

D

d.`BL sin ((theta)/(2)).(gL)^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • ELECTROMAGNETIC INDUCTION

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION - B)|22 Videos

  • ELECTROMAGNETIC INDUCTION

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION - C)|10 Videos

  • ELECTROMAGNETIC INDUCTION

    AAKASH INSTITUTE ENGLISH|Exercise Try Yourself|42 Videos

  • ELECTRIC CHARGES AND FIELDS

    AAKASH INSTITUTE ENGLISH|Exercise comprehension|3 Videos

  • ELECTROMAGNETIC WAVES

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION - D Assertion-Reason Type Questions|25 Videos

Similar Questions

Explore conceptually related problems

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle 2 theta . The earth's magnetic field component in the direction perpendicular to swing is B . Maximum potential difference induced across the pendulum is

A simple pendulum with a bob of mass m and a conducting wire of length L, swings under gravity with an angular amplitude 2theta . If the horizontal component of the earth's magnetic field perpendicular to the plane of motion of the pendulum is B, then the maximum emf induced across the pendulum is

A conducting rod AB of length l=1m is moving at a velocity v_A=4m//s making an angle 30^@ with its length. A uniform magnetic field B=2T exists in a direction perpendicular to the plane of motion. Then

A conducting rod PQ of length l = 2m is moving at a speed of 2m s^(-1) making an angle of 30^(@) with its length. A uniform magnetic field B = 2T exists in a direction perpendicular to the plane of motin. Then

A conducting rod PQ of length l = 2m is moving at a speed of 2m s^(-1) making an angle of 30^(@) with its length. A uniform magnetic field B = 2T exists in a direction perpendicular to the plane of motin. Then

A simple pendulum of length 1 m with a bob of mass m swings with an angular amplitude 30^(@) . Then (g= 9.8m//s^(2)

A simple pendulum has a bob of mass m and swings with an angular amplitude phi . The tension in thread is T . At a certain time the string makes an angle theta with the vertical (theta le phi)

A conducting rod AB of lemgth 2 l moves through a uniform magnetic field B. The rod makes an angle theta with the vertical while it falls with a uniform linear velocity v. The potential difference between the two ends of the rod is Bv^(2)l .

A uniform wire of resistance per unit length lambda is bent into semicircle of radius a . The wire rotates with angular velocity omega in a vertical plane about a orizontal is passing through C . A uniform magnetic field B exists in space in a direction perpendicular to paper inwards. a. Calculate potential difference between points A and D . which point is at higher potential? b. If points A and D are connected by a conducting wire of zero resistance, find the potential difference between A and C .

A straight rod of length L is rotating about an axis passing through O and perpendicular to the plane. In the space a uniform magnetic field B exits normal to the plane of rotation. Potential difference between a & b is equal to

AAKASH INSTITUTE ENGLISH-ELECTROMAGNETIC INDUCTION-Assignment (SECTION - A)
  1. A time varying magnetic flux passing through a coil is given by phi=xt...

    Text Solution

    |

  2. A metallic ring is dropped down, keeping its plane perpendicular to a ...

    Text Solution

    |

  3. Lights of a car become dim when the starter is operated. Why?

    Text Solution

    |

  4. Coefficient of coupling between two coils of self-inductances L(1) and...

    Text Solution

    |

  5. There is uniform magnetic field directed perpendicular and into the pl...

    Text Solution

    |

  6. A square loop of wire with side length 10 cm is placed at angle of 45^...

    Text Solution

    |

  7. Near a circular loop of conducing wires as shown, an electron moves al...

    Text Solution

    |

  8. A uniformly wound solenoid coil of self inductance 1.8xx10^(-4)H and r...

    Text Solution

    |

  9. The magnitude of the earth's magnetic field at a place is B(0) and an...

    Text Solution

    |

  10. Two parallel rails of a railway track insulated from each other and wi...

    Text Solution

    |

  11. When a low flying aircraft passes overhead, we sometimes notice a slig...

    Text Solution

    |

  12. A simple pendulum with bob of mass m and conducting wire of length L s...

    Text Solution

    |

  13. Electric field induced by changing magnetic fields are

    Text Solution

    |

  14. A circular loop of radius R, carrying current I, lies in XY-plane with...

    Text Solution

    |

  15. A helicopter rises vertically upwards with a speed of 100 m//s. If the...

    Text Solution

    |

  16. The energy stored in an inductor of self inductance L carrying current...

    Text Solution

    |

  17. An electric motor turns on DC source of emf 200 V and drawn a current ...

    Text Solution

    |

  18. Induction furnace make use of

    Text Solution

    |

  19. Two identical circular loops of metal wires are lying on a table witho...

    Text Solution

    |

  20. A short solenoid of length 4 cm, radius 2 cm and 100 turns is placed i...

    Text Solution

    |