Question 101:
easyA charge particle having mass \(m\) and charge \(q\) moving with speed \(v\) in uniform transverse magnetic field \(B\). The radial acceleration of charge particle will be
The magnetic force provides the necessary centripetal force: \(F = qvB = m a_r\). Thus, the radial acceleration is \(a_r = \frac{qvB}{m}\).
Question 102:
easyThe magnetic moment of a magnet having dimensions (10 cm × 5 cm × 4 cm) is \(2.5 \text{A m}^2\). The intensity of magnetisation of magnet is
Intensity of magnetisation \(I = \frac{M}{V}\). Volume \(V = 0.10 \times 0.05 \times 0.04 = 2 \times 10^{-4} \text{m}^3\). So, \(I = \frac{2.5}{2 \times 10^{-4}} = 1.25 \times 10^4 \text{A/m}\).
Question 103:
easyTwo long thin parallel wires separated by a distance \(d\) are carrying a current \(i\) each. The magnitude of the force per unit length applied by one wire on the other is
The magnetic force per unit length between two parallel current-carrying wires is given by \(F/L = \frac{\mu_0 i_1 i_2}{2\pi d}\). Since \(i_1 = i_2 = i\), \(F/L = \frac{\mu_0 i^2}{2\pi d}\), which is directly proportional to \(i^2\).
Question 104:
easyA charged particle is moving on circular path with velocity \(v\) in a uniform magnetic field \(B\). If velocity of the particle and strength of magnetic field is doubled, then time taken to complete one revolution becomes
The time period of revolution in a magnetic field is \[T = \frac{2\pi m}{qB}\]. It is independent of the velocity \(v\) and inversely proportional to \(B\). Thus, doubling \(B\) makes the time period half of its initial value.
Question 105:
easyA material suitable for making electromagnets should have
An electromagnet should become strongly magnetized when current is on, and lose its magnetization quickly when current is off. Therefore, it requires high retentivity and low coercivity.
Question 106:
easyA charged particle moves in a circular path under the action of uniform transverse magnetic field. Which of the given statement should be correct based on above information?
In a uniform magnetic field, the magnetic force is always perpendicular to velocity. Hence, work done is zero and kinetic energy remains constant. However, direction of velocity changes continuously, so momentum changes.
Question 107:
easyConsider the following statements:
Statement (A): All magnetism is due to circulating currents i.e. there cannot be magnetic field without current.
Statement (B): The orbital magnetic moment of an electron in the ground state of a hydrogen atom is equal to one Bohr Magneton.
Choose the correct option.
Statement (A) is incorrect because elementary particles like electrons have intrinsic spin magnetic moments not due to actual circulating currents. Statement (B) is correct as \(\mu_l = 1 \mu_B\) for ground state.
Question 108:
easyTesla is the unit of
(1997, 88)
The SI unit of magnetic field (magnetic induction) is Tesla ($ \text{T} $), defined as one Weber per square meter ($ \text{Wb/m}^2 $).
Question 109:
easyThe magnetic field at a distance r from a long wire carrying current i is 0.4 tesla. The magnetic field at a distance 2r is
(1992)
Magnetic field due to a long straight wire is inversely proportional to distance ($B \propto 1/r$). When distance is doubled from $r$ to $2r$, the magnetic field is halved. Therefore, the new magnetic field is $0.4 / 2 = 0.2\text{ tesla}$.
Question 110:
easyA long solenoid of radius $1\text{ mm}$ has $100$ turns per mm. If $1\text{ A}$ current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
(2022)
The magnetic field inside a long solenoid is given by $B = \mu_0 n I$. Given $n = 100\text{ turns/mm} = 10^5\text{ turns/m}$ and $I = 1\text{ A}$, substituting the values yields $B = (4\pi \times 10^{-7}) \times 10^5 \times 1 = 12.56 \times 10^{-2}\text{ T}$.