Two identical bar magnets are joined to form a cross.If this combination is suspended freely in a uniform field the angle made by the magnets with fields direction are respectively:

60^{o}, 30^{o}

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37^{o}, 53^{o}

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45^{o}, 45^{o}

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20^{o}, 70^{o}

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Solution

The correct option is C $45^{o}, 45^{o}$
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Similar questions

Two magnets of magnetic moments $M$ and $\sqrt{3} M$ are joined to form a cross $(+)$ . The combination is suspended freely in a uniform magnetic field. In equilibrium position the angle between the magnetic moment M and the field is:

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A bar magnet of length $10 c m$ experiences a torque of $\frac{\sqrt{2}}{10} N m$ in an uniform magnetic field of induction $0.4 W b / m^{2}$ when it is suspended making an angle $45^{o}$ with the field. The pole strength of the magnet is ( in $A m$ ) :

Q. When a bar magnet is suspended in a uniform magnetic field, the torque acting on it will be :

a) maximum	e) $θ = 45^{o}$ with the field
b) half the maximum field	f) $θ = 60^{o}$ with the field
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Two identical bar magnets are joined to form a cross.If this combination is suspended freely in a uniform field the angle made by the magnets with fields direction are respectively:

The correct option is C 45o,45oundefined

The correct option is C $45^{o}, 45^{o}$
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