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Tangent galvanometer is placed at a loca...

Tangent galvanometer is placed at a location where horizontal component of earth's magnetic field is `B_H = 3.0 xx 10^(-5)` T. Radius of the coil used in galvanometer is 25 cm and number of turns are 100. Find the current flowing through the galvanometer if deflection is `60^@`

Text Solution

AI Generated Solution

To find the current flowing through the tangent galvanometer, we can use the relationship between the magnetic field produced by the coil and the horizontal component of the Earth's magnetic field. The formula we will use is derived from the tangent law, which states: \[ B = B_H \tan(\theta) \] Where: - \( B \) is the magnetic field due to the coil, ...
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A tangent galvanometer has a coil of 50 turns and a radius of 20cm. The horizontal component of the earth's magnetic field is B_H = 3 xx 10^(-5) T . Find the current which gives a diflection of 45^) .

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Knowledge Check

  • At a given place on the earth's surface, the horizontal component of earth's magnetic field is 3 xx 10^(-5) T and resultant magnetic field is 6 xx 10^(-5) T. The angle of dip at this place is

    A
    `30^(@)`
    B
    `40^(@)`
    C
    `50^(@)`
    D
    `60^(@)`

  • A tangent galvanometer has a coil of 25 turns and radius of 15 cm. The horizontal component of the earth's magnetic field is 3xx10^(-5)T . The current required to producea defection of 45^(@) in it, is

    A
    `0.29A`
    B
    `1.2A`
    C
    `3.6xx10^(-5)A`
    D
    `0.14A`

  • The magnetic needle of a tangent galvanometer is deflected at an angle 30^(@) due to a magnet. The hoeizontal component of earth's magnetic field 0.34xx10^(-4)T is along the plane of the coil. The magnetic intensity is

    A
    `1.96xx10^(-4)T`
    B
    `1.96xx10^(-5)T`
    C
    `1.96xx10^(4)T`
    D
    `1.96xx10^(5)T`

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    The horizontal component of the earth's magnetic field is 3.6 xx 10^(-5) T where the dip is 60^@ . Find the magnitude of the earth's magnetic field.

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