# Torque Acting on a Bar Magnet Suspended in Magnetic Field

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#### Example – 01:

• A torque of moment 25 x 10-2 Nm acts on a magnet suspended in a uniform magnetic field of induction 0.5 Wb/m2 when making an angle of 30° with the field. Find the magnetic dipole moment of the magnet.
• Given: Torque = τ = 25 x 10-2 Nm, Magnetic induction = B = 0.5 Wb/m2, angle with field = θ = 30°.
• To find: Magnetic moment = M = ?
• Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 25 x 10-2/ (0.5 x sin 30°) = 25 x 10-2/ (0.5 x 0.5)

∴ M = 25 x 10-2/ (0.25) = 1 Am2

Ans: The magnetic dipole moment of the magnet is 1 Am2

#### Example – 02:

• A magnet of magnetic dipole moment 2 Am2 is deflected through 30° from the direction of a magnetic field of induction 2 Wb/m2. Find the magnitude of the torque or couple.
• Given: Magnetic moment = M = 2 Am2, Magnetic induction = B = 2 Wb/m2, angle with field = θ = 30°.
• To find: Torque = τ =?
• Solution:

τ = MB sin θ

∴ τ = 2 x 2 x sin 30° = 2 x 2 x 0.5 = 2 Nm

Ans: The torque acting on magnet is 2 Nm

#### Example – 03:

• A bar magnet of dipole moment 7.5 Am2 experiences a torque of 1.5 x 10-4 Nm, when placed inclined at 30° in a uniform magnetic field. Find the magnetic induction of the field.
• Given: Magnetic moment = M = 7.5 Am2, , angle with field = θ = 30°, Torque = τ = 1.5 x 10-4 Nm
• To find: Magnetic induction = B =?
• Solution:

τ = MB sin θ

∴ B = τ/ Msin θ

∴ B = 1.5 x 10-4/ (7.5 x sin 30°) = 1.5 x 10-4/ (7.5 x 0.5)

∴ B = 4 x 10-5 Wb/m2

Ans: The magnetic induction is  is 4 x 10-5 Wb/m or 4 x 10-5 T

#### Example – 04:

• A magnet of dipole moment 0.05 Am2 is suspended so to move freely in a horizontal plane. Find the couple required to hold it at right angles to the Earth’s horizontal magnetic induction of 0.32 x 10-4 Wb/m2.
• Given: Magnetic moment = M = 0.05 Am2, Magnetic induction = B = 0.32 x 10-4 Wb/m2, angle with field = θ = 90°.
• To find: Torque = τ =?
• Solution:

τ = MB sin θ

∴ τ = 0.05 x 0.32 x 10-4 x sin 90° = 0.05 x 0.32 x 10-4 x 1 = 1.6 x 10-6 Nm

Ans: The torque acting on magnet is 1.6 x 10-6 Nm

#### Example – 05:

• Calculate the dipole moment of magnet which when placed at right angles to the earth’s horizontal magnetic induction 2 x 10-5 Wb/m2 experiences a couple of 2 x 10-5 Nm.
• Given: Torque = τ = 2 x 10-5 Nm, Magnetic induction = B = 2 x 10-5  Wb/m2, angle with field = θ = 90°.
• To find: Magnetic moment = M = ?
• Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 2 x 10-5/ (2 x 10-5 x sin 90°) = 2 x 10-5/ (2 x 10-5 x 1)

∴ M = 1 Am2

Ans: The magnetic moment of the magnet is 1 Am2

#### Example – 06:

• A magnet of moment 0.6 Am2 makes an angle of 30° with the magnetic meridian. Calculate the torque that tends to bring  the magnet back to the meridian given that the earth’s horizontal field is 3.2 x 10-5 Wb/m2.
• Given: Magnetic moment = M = 0.6 Am2, Magnetic induction = B = 3.2 x 10-5 Wb/m2, angle with field = θ = 90°.
• To find: Torque = τ =?
• Solution:

τ = MB sin θ

∴ τ = 0.6 x 3.2 x 10-5 x sin 30° = 0.6 x 3.2 x 10-5 x 0.5 = 9.6 x 10-6 Nm

Ans: The torque required is 9.6 x 10-6 Nm

#### Example – 07:

• A torque of 5 x 10-3  Nm is required to hold a freely suspended horizontal bar magnet with the axis at an angle of 60° to uniform horizontal field of induction 3 x 10-3 Wb/m2. Find the magnetic moment of the magnet.
• Given: Torque = τ = 5 x 10-3 Nm, Magnetic induction = B = 3 x 10-3  Wb/m2, angle with field = θ = 60°.
• To find: Magnetic moment = M = ?
• Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 5 x 10-3/ (3 x 10-3 x sin 60°) = 5 x 10-3/ (3 x 10-3 x 0.8660)

∴ M = 1.925 Am2

Ans: The magnetic moment of the magnet is 1.925 Am2

#### Example – 08:

• A short bar magnet placed with its axis at 30° to a uniform magnetic field of 0.2 T experiences a torque of 0.06 Nm. Calculate the magnetic moment of the magnet and find out what orientation of the magnet corresponds to stable equilibrium in magnetic field.
• Given: Torque = τ = 0.06 Nm, Magnetic induction =0.2 T, angle with field = θ = 30°.
• To find: Magnetic moment = M = ? θ = ? for stable equilibrium.
• Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 0.06/ (0.2 x sin 30°) = 0.06/ (0.2 x 0.5)

∴ M = 0.6 Am2

For stable equilibrium torque acting on magnet is zero.

τ = MB sin θ

∴  0 = MB sin θ

∴  sin θ = 0

∴  θ = 0°

Ans: The magnetic moment of the magnet is 0.6 Amand for stable equilibrium θ = 0°

i.e. the axis of magnet should be parallel to the magnetic field.

#### Example – 09:

• A bar magnet of moment  0.9 JT-1 experiences a torque of 0.063 Nm, when placed inclined at 30° in a uniform magnetic field. Find the magnetic induction of the field.
• Given: Magnetic moment = M = 0.9 JT-1, angle with field = θ = 30°, Torque = τ = 0.063 Nm
• To find: Magnetic induction = B =?
• Solution:

τ = MB sin θ

∴ B = τ/ Msin θ

∴ B = 0.063/ (0.9 x sin 30°) = 0.063/ (0.9 x 0.5)

∴ B = 0.14 T

Ans: The magnetic induction is  is 0.14 T.

#### Example – 10:

• A magnet of dipole moment 6.4 Am2 and length 5 x 10-3 m makes an angle of 60° with the magnetic field of 0.4 T. calculate the torque acting on it. Also find the pole strength of the magnet.
• Given: Magnetic moment = M = 6.4 Am2, Magnetic induction = B = 0.4 T, angle with field = θ = 60°, magnetic length = 5 x 10-3 m
• To find: Torque = τ =? pole strength = m = ?
• Solution:

τ = MB sin θ

∴ τ = 6.4 x 0.4 x sin 60° = 6.4 x 0.4 x 0.8860 = 2.2 Nm

M = m x magnetic length

∴  m = M/magnetic length = 6.4/5 x 10-3 = 1.28 x 103 Am

Ans: The torque required is 1.28 x 103 Nm

#### Example – 11:

• Two magnets of magnetic moments M and 3M are joined together to form a cross. The combination is suspended freely in a uniform magnetic field. In the equilibrium position the magnet of magnetic moment M makes an angle θ with the field. Determine θ.
• Solution:

The magnet of magnetic moment M makes an angle θ with the field.

As the two magnets are forming cross, the magnet of magnetic moment 3M makes an angle (90° – θ) with the field.

Let B be the magnetic strength of the external field.

For equilibrium the torque acting on the two magnets must be equal.

τ1 = τ2

∴  MB sin θ =  3MB sin (90° – θ)

∴  sin θ =  3 cos θ

∴  tan θ =  3

∴  θ =  60°

Ans: Hence the value of θ is  60°

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