## Magnetic Effect Of Current And Magnetism

## Magnetic Properties Of Materials Long Questions and Answers

**Question 1. Indicate the magnetic axis and magnetic length for a permanent horse-shoe (U-shaped) magnet.**

**Answer:**

In the straight line CD passes through the two poles of the horseshoe magnet. So, CD is the magnetic axis.

The line segment AB indicates the least distance between the two poles, N and S.

So, the length of AB is the magnetic length.

**Question 2. A magnetized steel wire is bent in the form of L. One arm of E is of length 4 cm and the other arm Is of length 3 cm. If the magnetic moment before bending is pm, what will be the new magnetic moment?**

**Answer:**

Length of the steel wire before bending,

l = 4 + 3

= 7 cm

∴ Pole-strength, \(m=\frac{p_m}{l}\)

Distance between the two poles after bending,

⇒ \(l^{\prime}=\sqrt{4^2+3^2}=5 \mathrm{~cm}\)

So, the new magnetic moment,

⇒ \(p_m^{\prime}=m \cdot l^{\prime}=\frac{p_m}{l} \cdot l^{\prime}=\frac{p_m}{7} \cdot 5=0.714 p_m\)

**Question 3. Two identical bar magnets having magnetic moments 2 pm and 3 pm are kept one over the other in such a manner that their**

**Like poles****Opposite poles are in contact. Determine the resultant magnetic moments in each case**

**Answer:**

1. Magnetic moment is a vector quantity. If the like poles are in contact, the magnetic moments for the two magnets are in the same direction. So, the resultant magnetic moment = 2 p_{m} + 3 p_{m} = 5 pm, and this resultant magnetic moment acts in the direction of 3 pm and 2 pm.

2. If the opposite poles are in contact, 3 p_{m} and 2 p_{m} lie on the same straight line but opposite in direction. So, the resultant magnetic moment = 3 p_{m} – 2 p_{m} = p_{m}, and this resultant magnetic moment acts in the direction of 3 p_{m}.

**Question 4. If a permanent bar magnet is cut along its breadth into two equal parts, what will be the pole strength and magnetic moment of each part?**

**Answer:**

Let the pole strength of the bar magnet be m and its magnetic length is 2l.

∴ The magnetic moment of the bar magnet,

Pm = m.2l

If the magnet is cut along its breadth into two equal parts, the strength of each part remains m but the magnetic length of each part becomes l.

So, the magnetic moment of each part,

⇒ \(p_m^{\prime}=m \cdot l=\frac{p_m}{2}\)

**Question 5. If a permanent magnet is cut along the length into two equal parts, what will be the pole strength and magnetic moment of each part?**

**Answer:**

Let the pole strength of the bar magnet be m and its magnetic length is 2l.

So, the magnetic moment of the magnet, p_{m} = m.2l.

If the magnet is cut along its length into two equal parts, the number of free poles of the molecular magnets at its extremities will also be halved.

So, the poie-strength of each part = \(\frac{m}{2}\) and magnetic length = 2l.

∴ Magnetic moment, \(p_m^{\prime}=\frac{m}{2} \cdot 2 l=m l=\frac{p_m}{2}\)

**Question 6. What is a ‘magnet-proof’ watch?**

**Answer:**

A ring of soft iron is fitted around a watch to make it free from external magnetic influence. That ring acts as a magnetic screen. This kind of watch is called a ‘magnet-proof’ watch.

**Question 7. An electron (charge = e) revolves around a nucleus along a circular path of radius r with frequency f. What will be the magnetic moment of the electron due to its orbital motion?**

**Answer:**

Number of complete revolutions per second = f.

So, the amount of charge flowing through any point on the orbit per second = effective current (I) = ef.

Therefore, the required magnetic moment,

⇒ \(p_m=I A=e f \cdot \pi r^2\)

**Question 8. According to the mariner’s compass, a ship is sailing toward the east. If the declination at the place is 20°E, what is the actual direction of motion of the ship?**

**Answer:**

Let the ship be at A. The points E and E’ are on the geographical east and magnetic east of the point A.

According to the problem,

⇒ \(\angle N A N^{\prime} \doteq 20^{\circ}\)

∴ \(\angle E A E^{\prime}=20^{\circ}\)

So, the ship is moving towards point E’, i.e., along a direction of 20° south of the geographically east

**Question 9. Example why a mariner’s compass does not work inside a submarine?**

**Answer:**

Due to the presence of a geomagnetic field, a mariner’s compass indicates directions. But the iron covering of a submarine acts as a magnetic screen. As a result, geomagnetic fields cannot penetrate the interior of a submarine. Thus, a mariner’s compass does not work inside a submarine.

**Question 10. In a hydrogen atom, an electron of charge e revolves in an orbit of radius r with speed v. Find the magnetic moment associated with the electron.**

**Answer:**

The period of revolution of the electron, \(T=\frac{2 \pi r}{v}\)

equivalent current, \(I=\frac{e}{T}=\frac{e v}{2 \pi r}\)

∴The magnetic moment of the electron,

⇒ \(p_m=I A=\frac{e v}{2 \pi r} \cdot \pi r^2=\frac{1}{2} e v r\)

**Question 11. A vertical iron pillar, partially dipped inside the ground, is found to be magnetized after several years. What will be the polarity at the top of the pillar when it is in the northern hemisphere of the earth?**

**Answer:**

The vertical iron pillar gets magnetic induction in the presence of the earth’s magnetic field. We know that the south pole of Earth’s magnet lies in the northern hemisphere. As a result, following the rule of magnetic induction, a north pole will be induced at the underground end, and a south pole at the top of the pillar.

**Question 12. A magnetic needle lying parallel to a magnetic field requires a W unit of work to turn it through 60°. How much torque should be needed to maintain the needle in this position?**

**Answer:**

Work done in rotating the magnetic needle from 0° position through an angle θ in a magnetic field B is,

W = pmB(l- cosθ) [where pm = magnetic moment of the needle]

⇒ \(p_m^B=\frac{W}{1-\cos \theta}\)

∴ Required torque,

⇒ \(\tau=p_m^B \sin \theta=\frac{W \sin \theta}{1-\cos \theta}=\frac{W \sin 60^{\circ}}{1-\cos 60^{\circ}}=\sqrt{3} W\)

**Question 13. I ampere current is flowing through a meter-long conducting wire. If the wire is shaped into a circular loop, then what will be its magnetic moment?**

**Answer:**

If the radius of the circular loop is r then,

⇒ \(2 \pi r=l. \quad \text { or, } r=\frac{l}{2 \pi}\)

and area of the loop, \(A=\pi r^2=\frac{l^2}{4 \pi}\)

∴ The magnetic moment of the circular loop = \(I A=\frac{l^2}{4 \pi}\)

**Question 14. Indicate the zero-potential line of a bar magnet.**

**Answer:**

The perpendicular bisector of the magnetic axis of a bar magnet is the zero-potential line because any point lying on this line is equidistant from the two poles of the magnet. Let O be any point on the perpendicular bisector of the magnetic axis of the magnet NS having pole-strength q_{m}. If the distance of the point O from each pole is r, then the potential at O is,

⇒ \(V=\frac{q_m}{r}-\frac{q_m}{r}=0\)

**Question 15. Two particles, each of mass m and charge q, are kept at the two ends of a light rod of length 2l and are rotated with a uniform angular velocity about the vertical axis passing through the center of the rod. Determine the ratio of the magnetic moment and the angular momentum of the combination with respect to the center of the rod.**

**Answer:**

The radius of the path described = l.

If the angular velocity = ω, time period, \(T=\frac{2 \pi}{\omega}\)

In this duration of time, the amount of charge flowing through any point of the orbit = 2q.

So, the charge flows per second

= effective current (I)

⇒ \(\frac{2 q}{T}=\frac{q \omega}{\pi}\)

∴ Magnetic moment, \(p_m=L A=\frac{q \omega}{\pi} \cdot \pi l^2=q \omega l^2\)

and angular momentum, \(L=(2 m) l^2 \omega=2 m \grave{\omega} l^2\)

Therefore, \(\frac{p_m}{L}=\frac{q}{2 m}\)

**Question 16. An iron nail gains kinetic energy due to the force of attraction by a magnet. What is the source of this kinetic energy?**

**Answer:**

When an iron nail is kept in a magnetic field, it gains some magnetic potential energy. In this case, that potential energy is the source of the kinetic energy of the nail. When the nail. moves towards the magnet due to the force of attraction, the potential energy of the nail decreases and it is converted into the said kinetic energy.