$A$ tangent galvanometer has a coil of $25$ turns and a radius of $15\, cm$. The horizontal component of the earth's magnetic field is $3 \times 10^{-5}\, T$. The current required to produce a deflection of $45^{\circ}$ in it is....$A$

  • A
    $0.29$
  • B
    $1.2$
  • C
    $3.6 \times 10^{-5}$
  • D
    $0.14$

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Similar Questions

$A$ tangent galvanometer shows a deflection of $45^{\circ}$ when a $10 \, mA$ current passes through it. If the horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \, T$ and the radius of the coil is $10 \, cm$,find the number of turns in the coil.

$A$ vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \, s$ in the earth's horizontal magnetic field of $24 \, \mu T$. When a horizontal field of $18 \, \mu T$ is produced opposite to the earth's field by placing a current-carrying wire,the new time period of the magnet will be....$s$

$A$ bar magnet $A$ of magnetic moment $M_A$ is found to oscillate at a frequency twice that of magnet $B$ of magnetic moment $M_B$ when placed in a vibrating magnetometer. We may say that

$A$ short bar magnet having a magnetic moment $4 \text{ Am}^2$,placed in a vibrating magnetometer,vibrates with a time period of $8 \text{ s}$. Another short bar magnet having a magnetic moment $8 \text{ Am}^2$ vibrates with a time period of $6 \text{ s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \text{ kg-m}^2$,the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field).

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ (about its center,perpendicular to its length) is cut into two equal pieces,perpendicular to its length. Let $T$ be the period of oscillation of the original magnet about an axis through its midpoint,perpendicular to its length,in a uniform magnetic field $B$. What would be the period $T'$ for each piece?

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