The magnetic moment produced in a substance of \(1\text{ gm}\) is \(6 \times 10^{-7}\text{ A-m}^2\). If its density is \(5\text{ gm/cm}^3\), then the intensity of magnetisation in \(A/m\) will be:
1. \(8.3 \times 10^6\)
2. \(3.0\)
3. \(1.2 \times 10^{-7}\)
4. \(3 \times 10^{-6}\)
View Answer
Intensity of magnetisation is \(I = \frac{M}{V} = \frac{M\rho}{m}\). Given \(m = 1\text{ gm}\), \(M = 6 \times 10^{-7}\text{ A-m}^2\), \(\rho = 5 \times 10^3\text{ kg/m}^3\). Thus \(I = \frac{6 \times 10^{-7} \times 5 \times 10^3}{10^{-3}} = 3.0\text{ A/m}\).
If magnetic field in space is \(1\text{ T } \hat{i}\), electric field is \(10\text{ N/C } \hat{i}\), no gravitational field is present and a charged particle is released from rest from origin, it will:
1. not move at all
2. move in circular path
3. move in a helical path
4. move on a straight line
View Answer
Since the particle starts from rest, its initial magnetic force is zero. The electric field accelerates it along \(\hat{i}\). Because velocity remains parallel to the magnetic field, the magnetic force remains zero, and it continues on a straight line.
Statement-1: In an isolated conductor, free electrons keep on moving but no net magnetic force acts on a conductor in a magnetic field.
Statement-2: In a conductor, the average velocity of thermal motion of electrons is zero. Hence no current flows through the conductor.
1. Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.
2. Both Statement-1 and Statement-2 are true but Statement-2 is not correct explanation of Statement-1.
3. Statement-1 is true but Statement-2 is false.
4. Statement-1 and Statement-2 are false.
View Answer
The net magnetic force on a current-carrying conductor is given by \(F = I L B\). Since average velocity of thermal motion is zero, current \(I = 0\), resulting in zero net force.
A magnetic field strength (\(H\)) equal to \(3 \times 10^3\text{ A m}^{-1}\) produces a magnetic field of induction (\(B\)) equal to \(12\pi\) tesla in an iron rod. The relative permeability of the iron rod is
1. \(10^3\)
2. \(10^2\)
3. \(10^4\)
4. \(10^5\)
View Answer
The relationship is \(B = \mu H = \mu_r \mu_0 H\). Given \(B = 12\pi\text{ T}\), \(H = 3 \times 10^3\text{ A m}^{-1}\), and \(\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}\). Thus, \(12\pi = \mu_r (4\pi \times 10^{-7})(3 \times 10^3)\), which simplifies to \(\mu_r = 10^4\).
If the direction of the initial velocity of a charged particle is neither along nor perpendicular to a uniform magnetic field, then the path of charged particle will be
1. An ellipse
2. A circle
3. A straight line
4. A helix
View Answer
When velocity vector is at an angle \(\theta\) (where \(0^\circ < \theta < 90^\circ\)) to the magnetic field, the component parallel to the field produces linear translation, while the perpendicular component produces circular motion. The combined path is a helix.
A long straight wire carries an electric current \(4\text{ A}\). The magnetic induction at a perpendicular distance \(2\text{ m}\) from the wire is
1. \(2 \times 10^{-7}\text{ T}\)
2. \(4 \times 10^{-7}\text{ T}\)
3. \(2 \times 10^{-8}\text{ T}\)
4. \(4 \times 10^{-8}\text{ T}\)
View Answer
The magnetic field near a long straight wire is given by \(B = \frac{\mu_0 I}{2\pi r}\). Substituting \(I = 4\text{ A}\) and \(r = 2\text{ m}\) with \(\mu_0 = 4\pi \times 10^{-7}\text{ T m/A}\) gives \(B = \frac{4\pi \times 10^{-7} \times 4}{4\pi} = 4 \times 10^{-7}\text{ T}\).
If a proton has velocity \((2\hat{i} + 3\hat{k})\) m/s and it is subjected to a magnetic field of \(4\hat{i}\) T, then its:
1. Speed will not change
2. Path will not change
3. Velocity will remain same
4. Momentum will remain same
View Answer
Since the magnetic force \(\vec{F} = q(\vec{v} \times \vec{B})\) is always perpendicular to the velocity, the work done is zero. Hence, the kinetic energy and speed remain constant.
A circular coil of radius \(R\) having current \(I\) is placed in a uniform magnetic field \(B\). If the angle between the area vector of the coil and the magnetic field is \(60^circ\), then the torque on the coil will be:
1. \[\frac{\pi R^2 I B}{2}\]
2. \[\frac{\sqrt{3}\pi R^2 I B}{2}\]
3. \(\pi R^2 I B\)
4. Zero
View Answer
The torque is given by \(\tau = MBsin\theta\), where \(M = I A = I(\pi R^2)\) and \(\theta = 60^\circ\). Thus, \(tau = I(\pi R^2)Bsin 60^\circ = \frac{\sqrt{3}\pi R^2 I B}{2}\).
A bar magnet has length \(3\text{ cm}\), cross-sectional area \(2\text{ cm}^2\) and magnetic moment \(3\text{ A m}^2\). The intensity of magnetisation of bar magnet is
1. \(2 \times 10^5\text{ A/m}\)
2. \(3 \times 10^5\text{ A/m}\)
3. \(4 \times 10^5\text{ A/m}\)
4. \(5 \times 10^5\text{ A/m}\)
View Answer
Intensity of magnetisation \(I = \frac{M}{V}\). Here, volume \(V = A \times L = (2 \times 10^{-4}\text{ m}^2) \times (3 \times 10^{-2}\text{ m}) = 6 \times 10^{-6}\text{ m}^3\). Thus, \(I = \frac{3}{6 \times 10^{-6}} = 5 \times 10^5\text{ A/m}\).