Question 131:
easyA wire of length $L\text{ m}$ carrying a current of $I\text{ A}$ is bent in the form of a circle. Its magnetic moment is:
(2020-Covid)
Radius $r = L / (2\pi)$. Area $A = \pi r^2 = L^2 / (4\pi)$. Magnetic moment $M = I A = I L^2 / (4\pi)$.
Question 132:
easyA charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given by:
(2007)
Equivalent current $I = q / T = qv / (2\pi R)$. Magnetic moment $\mu = I A = \frac{qv}{2\pi R} \pi R^2 = \frac{qvR}{2}$.
Question 133:
easyIf number of turn, area and current through it is given by $n$, $A$ and $i$ respectively then its magnetic moment will be:
(2001)
Magnetic dipole moment of a current-carrying coil is the product of number of turns, current, and area, i.e., $M = niA$.
Question 134:
easy5. A closely wound solenoid of 2000 turns and area of cross section $1.5 \times 10^{-4}\text{ m}^2$ carries a current of $2.0\text{ A}$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2}\text{ tesla}$ making an angle of $30^\circ$ with the axis of the solenoid. The torque on the solenoid will be (2010 Mains)
Magnetic dipole moment $M = NIA = 2000 \times 2.0 \times 1.5 \times 10^{-4} = 0.6\text{ A m}^2$.
Torque $\tau = MB \sin\theta = 0.6 \times (5 \times 10^{-2}) \times \sin 30^\circ = 1.5 \times 10^{-2}\text{ Nm}$.
Question 135:
easy6. A 250 turn rectangular coil of length $2.1\text{ cm}$ and width $1.25\text{ cm}$ carries a current of $85\text{ }\mu\text{A}$ and subjected to a magnetic field of strength $0.85\text{ T}$. Work done for rotating the coil by $180^\circ$ against the torque is: (2017-Delhi)
Magnetic moment $M = NIA = 250 \times (85 \times 10^{-6}\text{ A}) \times (2.1 \times 1.25 \times 10^{-4}\text{ m}^2) \approx 5.58 \times 10^{-6}\text{ A m}^2$.
Work done $W = MB(1 - \cos 180^\circ) = 2MB = 2 \times (5.58 \times 10^{-6}) \times 0.85 \approx 9.1\text{ }\mu\text{J}$.
Question 136:
easy4. A bar magnet of magnetic moment $M$ is cut into two parts of equal length. The magnetic moment of each part will be (1997)
When a bar magnet is cut into two equal parts perpendicular to its length, the length of each piece becomes $L/2$ while the pole strength $m$ remains unchanged.
New magnetic moment $M' = m \times (L/2) = M/2 = 0.5M$.
Question 137:
easy7. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now the torque required to keep the magnet in this new position is: (2016 – II)
Work done, $W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} \Rightarrow MB = 2W$.
Torque required, $\tau = MB \sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3}W$.
Question 138:
easy8. A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be: (2012 Mains)
$W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} = \sqrt{3} \Rightarrow MB = 2\sqrt{3}$ J.
Torque, $\tau = MB \sin 60^{\circ} = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$ J.
Question 139:
easy9. A short bar magnet of magnetic moment $0.4$ J T$^{-1}$ is placed in a uniform magnetic field of $0.16$ T. The magnet is in stable equilibrium when the potential energy is: (2011 Mains)
For stable equilibrium, the angle between $M$ and $B$ is $0^{\circ}$.
Potential energy $U = -MB \cos 0^{\circ} = - (0.4)(0.16)(1) = -0.064$ J.
Question 140:
easy12. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It: (2012 Pre)
At the geomagnetic poles, the Earth's magnetic field is entirely vertical, meaning the horizontal component $B_H$ is zero. A compass free to rotate only horizontally experiences no directing torque and will stay in any position.