Two short bar magnets have magnetic moments $1.2 \text{ Am}^2$ and $1.0 \text{ Am}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres,such that their north poles point towards the geographic south. They share a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. Calculate the resultant horizontal magnetic induction at the midpoint of the line joining their centres. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2$)

  • A
    $3.6 \times 10^{-5} \text{ T}$
  • B
    $1.84 \times 10^{-4} \text{ T}$
  • C
    $2.56 \times 10^{-4} \text{ T}$
  • D
    $5.8 \times 10^{-5} \text{ T}$

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Similar Questions

Two short bar magnets $A$ and $B$ (having magnetic moments $M_{1}$ and $M_{2}$ respectively) are kept one above the other with their magnetic axes perpendicular to each other. If their resultant magnetic field at a point on the axis of magnet $A$ is inclined at $45^{\circ}$ with the axis of magnet $A$,then the ratio of magnetic moments $\frac{M_{2}}{M_{1}}$ is $[\tan 45^{\circ} = 1]$.

The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

Deduce the expression for the potential energy of a bar magnet in a uniform magnetic field and discuss special cases.

$A$ bar magnet has a magnetic moment of $200 \text{ A m}^2$. The magnet is suspended in a magnetic field of $0.30 \text{ N A}^{-1} \text{ m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$ will be:

$A$ short bar magnet of magnetic moment $5.25 \times 10^{-2} \;J\, T^{-1}$ is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet,the resultant field is inclined at $45^{\circ}$ with the earth's field on
$(a)$ its normal bisector and
$(b)$ its axis.
Magnitude of the earth's field at the place is given to be $0.42 \;G$. Ignore the length of the magnet in comparison to the distances involved.

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