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### Problems on Inverse Square Law

#### Example – 01:

- The force between two magnetic poles in air is 9.604 mN. If one pole is ten times stronger than other, calculate the pole strength of each magnet. The distance between the poles of two magnets is 0.1 m.
**Given:**Force = F = 9.604 mN = 9.604 x 10^{-3}N, distance between poles = r = 0.1 m, Relation between pole strength m_{2}= 10 m_{1}, μ_{o}/4π = 10^{-7}Wb/Am.**To Find:**pole strengths = ?**Solution:**

m_{2} = 10 m_{1} = 10 x 9.8 = 98 Am

**Ans:** The pole strengths of magnets are 9.8 Am and 98 Am.

#### Example – 02:

- What is the force of repulsion between two magnetic poles of strengths 1.6 Am and 7.2 Am separated by a distance 0.06 m in vacuum?
**Given:**pole strengths m_{1}= 1.6 Am and m_{2}= 7.2 Am, distance between poles = r = 0.06 m, μ_{o}/4π = 10^{-7}Wb/Am.**To Find:**The force between poles = F = ?**Solution:**

**Ans:** Force Between Poles is 3.2 x 10^{-4} N

### Problems on Magnetic Dipole Moment of Bar Magnet:

#### Example – 03:

- A bar magnet of geometric length 18 cm has pole strength 100 Am. Find the magnetic dipole moment of the bar magnet.
**Given:**Geometric length of bar magnet = 18 cm = 0.18 m, Pole strength = m = 100 Am**To Find:**Magnetic dipole moment = M = ?**Solution:**

Magnetic length = (5/6) x Geometric length = (5/6) x 0.18 = 0.15 cm

M = m x magnetic length

∴ M = 100 x 0.15 = 15^{ }Am^{2}

**Ans:** The magnetic dipole moment of bar magnet is 15 Am^{2}.

#### Example – 04:

- A bar magnet of magnetic length 0.1 m has pole strength of 10 Am. What is dipole moment of this magnet.
**Given:**Magnetic length of bar magnet = 0.1 m, Pole strength = m = 10 Am**To Find:**Magnetic dipole moment = M = ?**Solution:**

M = m x magnetic length

∴ M = 10 x 0.1 = 1^{ }Am^{2}

**Ans:** The magnetic dipole moment of bar magnet is 1 Am^{2}.

#### Example – 05:

- A bar magnet of magnetic moment 5 Am
^{2}has poles 0.2 m apart. Calculate the pole strength. **Given:**Magnetic moment = 5 Am^{2}, magnetic length of bar magnet = 0.2 m,**To Find:**pole strength = m = ?**Solution:**

M = m x magnetic length

∴ m = M/ magnetic length = 5/0.2 = 25 Am

**Ans:** The pole strength of bar magnet is 25 Am.

**Example – 06:**

- A bar magnet has pole strength 48 Am which are 0.25 m apart. What is the magnetic moment of the magnet.
**Given:**Magnetic length of bar magnet = 0.25 m, Pole strength = m = 48 Am**To Find:**Magnetic dipole moment = M = ?**Solution:**

M = m x magnetic length

∴ M = 48 x 0.25 = 12^{ }Am^{2}

**Ans:** The magnetic dipole moment of bar magnet is 12 Am^{2}.

### Problems on Magnetic Dipole Moment of Current Carrying Coil:

**Example – 07:**

- A circular coil of 20 turns and radius 10 cm has a magnetic dipole moment of 3.142 Am
^{2}. What is the current flowing through the coil. **Given:**Number of turns = n = 20, radius of coil = 10 cm = 0. 1 m, magnetic dipole moment = M = 3.142 Am^{2}.**To Find:**Current through the coil = i = ?**Solution:**

**Ans:** The current through the coil is 5 A.

#### Example – 08:

- A circular coil of 50 turns and 10 cm radius carries a current of 2 A. Find the magnetic moment of the coil.
**Given:**Number of turns = n = 50, radius of coil = 10 cm = 0. 1 m, Current through coil = i = 2 A.**To Find:**Magnetic dipole moment = M = ?**Solution:**

M = n i A = n i (π r^{2})

∴ M = 50 x 2 x 3.142 x (0.1)^{2} = 3.142 Am^{2}

**Ans:** The magnetic dipole moment is 3.142 Am^{2}.

#### Example – 09:

- A circular coil has 1000 turns each of area 2 m
^{2}. If a current of 3 mA flows through the coil, find the magnetic moment of the coil. **Given:**Number of turns = n = 1000, Area of coil = A = 2 m^{2}, Current through coil = i = 3 mA = 3 x 10^{-3}A**To Find:**Magnetic dipole moment = M = ?**Solution:**

M = n i A

∴ M = 1000 x 3 x 10^{-3 }x 2 = 6^{ }Am^{2}

**Ans:** The magnetic dipole moment is 6 Am^{2}.

#### Example – 10:

- A rectangular coil of length 8 cm and breadth 5 cm has 200 turns of insulated wire. Find the magnetic dipole moment of the coil. When a current of 2 A flows through it.
**Given:**Number of turns = n = 200, Area of coil = A = 8 cm x 5 cm = 40 cm^{2}= 40 x 10^{-4}m^{2}, Current through coil = i = 2 A**To Find:**Magnetic dipole moment = M = ?**Solution:**

M = n i A

∴ M = 200 x 40 x 10^{-4 }x 2 = 1.6^{ }Am^{2}

**Ans:** The magnetic dipole moment is 1.6 Am^{2}.

#### Example – 11:

- A steel wire of length ‘
*l*‘ has a magnetic moment M. If the wire is bent in a semicircular arc, what would its moment be? **Solution:**

Let m be the pole strength

Now, M = m x magnetic length

∴ m = M/*l*

Curved length of semicircle = half circumference

∴ *l* = π r

∴ r = *l */π

Now new magnetic dipole moment

M = m x new magnetic length

∴ M = (M/*l*)x 2r = (M/*l*)x 2 x (*l */π) = 2M/π

Ans: New magnetic moment is 2M/π

Science > Physics > Magnetism >You are Here |

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