A circular loop carrying a current is replaced by ann equivalent magnetic dipole. A point on the axis of the loop is in:
- ✓End-on position.
- BBroadside-on position.
- CBoth.
- DNone.
Answer: A.
View full solution →Question types
Permanent Magnets question types
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Sample Questions
Permanent Magnets questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
A dip circle is taken to geomagnetic equator. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. The needle will stay:
- AIn horizontal direction only.
- BIn vertical direction only.
- CIn any direction except vertical and horizontal.
- ✓In the direction it is released.
Answer: D.
View full solution →A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the loop is in:
- AEnd-on position.
- ✓Broadside-on position.
- CBoth.
- DNone.
Answer: B.
View full solution →Two short magnets of equal dipole moments M are fastened perpendicularly at their centre $($figure$)$. The magnitude of the magnetic field at a distance d from the centre on the bisector of the right angle is:
- A$\frac{\mu_0}{4\pi}\frac{\text{m}}{\text{d}^3}$
- B$\frac{\mu_0}{4\pi}\frac{\sqrt{2}\text{M}}{\text{d}^3}$
- ✓$\frac{\mu_0}{4\pi}\frac{\sqrt[2]{2}\text{M}}{\text{d}^3}$
- D$\frac{\mu_0}{4\pi}\frac{2\text{M}}{\text{d}^3}$
Answer: C.
View full solution →Magnetic meridian is:
- AA point.
- BA line along north-south.
- CA horizontal plane.
- ✓A vertical plane.
Answer: D.
View full solution →Can we have a single north pole? A single south pole?
Do two distinct poles actually exist at two nearby points in a magnetic dipole?
Magnetic scalar potential is defined as$\text{U}(\overrightarrow{\text{r}_2})-\text{U}(\overrightarrow{\text{r}_1})=-\int\limits^{\vec{\text{r}}_2}_{\vec{\text{r}_1}} \vec{\text{B}}.\text{d}\vec{\text{l}}.$
Apply this equation to a closed curve enclosing a long atraicht wire. The RHS of the above equation is then $-{\mu}_\text{o} \text{ i}$ by Ampere's law. We see that $\text{U}(\vec{\text{r}_2})\neq\text{U}(\vec{\text{r}_1})$ even when $\vec{\text{r}_2}=\vec{\text{r}_1}.$Can we have a magnetic acalar potential in this case?
View full solution →Apply this equation to a closed curve enclosing a long atraicht wire. The RHS of the above equation is then $-{\mu}_\text{o} \text{ i}$ by Ampere's law. We see that $\text{U}(\vec{\text{r}_2})\neq\text{U}(\vec{\text{r}_1})$ even when $\vec{\text{r}_2}=\vec{\text{r}_1}.$Can we have a magnetic acalar potential in this case?
Can the dip at a place be zero? 90°?
Can the earth's magnetic field be vertical at a place? What will happen to a lreely suspended magnet at such a place? What is the value of dip here?
Two bar magnets are placed close to each other with their opposite poles facing each other. In absence of other force the magnets are pulled towards each other and their kinetic energy increases. Does it contradict our earher knowledge that magnetic forces cannot do any work and hence cannot increase kinetic energy of a system?
Compare the direction of the magnetic field inside a solenoid with that of the field there if the solenoid is replaced by its equivalent combination of north pole and south pole.
The force on a north pole,$\overrightarrow{\text{F}}=\text{m}\overrightarrow{\text{B}},$ is parallel to the field $\overrightarrow{\text{B}}.$ Does it contradict our earlier knowledge that a magnetic field can exert forces only perpendicular to itself?
View full solution →An iron needle is attracted to the ends of a bar magnet but not to the middle region of the magnet. Is the material making up the ends of a bar magnet different from that of the middle region?
Sketch the magnetic field lines for a current-carrying circular loop near its centre. Replace the loop by an equivalent magnetic dipole and sketch the magnetic field lines near the centre of the dipole. Identify the difference.
A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is $2\mu\text{T.}$ Another short magnet of magnetic moment $1.6A-m^2$ is placed 20cm east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north pole:
View full solution →- Towards north.
- Towards south.
The magnetic field due to the earth has a horizontal component of $26\mu\text{T}$ at a place where the dip is 60°. Find the vertical component and the magnitude of the field.
View full solution →A long bar magnet has a pole strength of 10A-m. Find the magnetic field at a point on the axis of the magnet at a distance of 5cm from the north pole of the magnet.
The needle of a dip circle shows an apparent dip of 45° in a particular position and 53° when the circle is rotated through 90°. Find the true dip.
Two long bar magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2·0cm from the south pole of the second. If both the magnets have a pole strength of 10A-m, find the force exerted by one magnet on the other.
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