The magnetic materials having negative magnetic susceptibility are:
Diamagnetic substances have negative magnetic susceptibility because they develop magnetization in a direction opposite to the applied magnetic field.
To increase the current sensitivity of a moving coil galvanometer, we should:
Current sensitivity is given by \(I_s = \frac{NBA}{C}\). To increase it, we need to increase \(N\), \(B\), \(A\) or decrease \(C\). None of the options perform these changes.
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:
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:
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.
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 long straight wire carries an electric current \(4\text{ A}\). The magnetic induction at a perpendicular distance \(2\text{ m}\) from the wire is
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 \(B = \frac{4\pi \times 10^{-7} \times 4}{4\pi} = 4 \times 10^{-7}\text{ T}\).
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
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
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.
In the product \(\vec{F} = q(\vec{v} \times \vec{B}) = q \vec{v} \times (B_x \hat{i} + B_y \hat{j} + B_0 \hat{k})\), for \(q = 1\) and \(\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}\) and \(\vec{F} = 4\hat{i} – 20\hat{j} + 12\hat{k}\). What will be the complete expression for \(vec{B}\)?
Using the relation \(\vec{F} = \vec{v} \times \vec{B}\), we compare vector components: \(4\hat{i} - 20\hat{j} + 12\hat{k} = (4 B_0 - 6 B_y)\hat{i} - (2 B_0 - 6 B_x)\hat{j} + (2 B_y - 4 B_x)\hat{k}\). Testing the values in Option C gives \(B_x = -6\, B_y = -6\, B_0 = -8\), which completely satisfies all equations.
A uniform bar magnet has magnetic moment \(M\). If it is cut into 4 equal parts parallel to its length such that the width of each part is same, then the magnetic moment of each part will be:
Cutting parallel to length into 4 equal parts keeps the length the same but reduces the pole strength to \(m' = frac{m}{4}\). Hence, the new magnetic moment is \(M' = m'L = \frac{M}{4}\).