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A tangent galvanometer ha s60 turns of w...

A tangent galvanometer ha s60 turns of wire of mean diameter 20 cm . Find the reduction factor of the tangent galvanometer at a place where the horizontal component of earth's magnetic field `0.32 xx 10^(-4) T`.

Text Solution

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The correct Answer is:
0.085 A
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A tangent galvanometer has 66 turns and the diameter of its coil is 22cm. It gives a deflection of 45^@ for 0.10 A current. What is the value of the horizontal component of the earth's magnetic field?

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Knowledge Check

  • A tangent gavanometer has 80 turns of wire. The internal and external diameters of the coil are 19 cm and 21 cm respectively. The reduction factor of the galvonometer at a place where H = 0.32 oersted will be (1 oersted =80 A//m)

    A
    `0.0064`
    B
    `0.64`
    C
    `0.064`
    D
    none of these

  • In a tangent galvanometer, the magnetic induction produced by the coil of wire situated in the magnetic meridian is found to be equal to the horizontal component of the earth's magnetic field. The deflection produced in it will be

    A
    `30^(@)`
    B
    `60^(@)`
    C
    `45^(@)`
    D
    `90^(@)`

    • Similar Questions

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    A tangent galvanometer has a cooil of 22 turns and mean radius 0.1 m. Find the reduction factor at a place , where B_(H)=0.3xx10^(-4)T .

    A coil of radius 10 cm and resistance 40 Omega has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180^@ . Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is B_H = 3.0X 10^(-5) T .

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    The coil in a tangent galvanometer is 16cm in radius. How many turns of the wire should be wound on it if a current of 20mA is to produce a deflection of 45^@ ? Take horizontal component of earth's field =0*36xx10^-4T .

    The radius of the coil of a tangent galvanometer is 16cm . How many turns of the wire should be used if a current of 40mA is to produce a deflection of 45^@ . Given, horizotnal component of earth's field is 0*36xx10^-4T.

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