Question 121:
easyA uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron:
(2011 Pre)
The magnetic force is zero because velocity and magnetic field are parallel. The electric force acts opposite to the electron's motion since it is negatively charged, causing its speed to decrease.
Question 122:
easyA positively charged particle moving due East enters a region of uniform magnetic field directed vertically upwards. This particle will
(1997)
Magnetic force is always perpendicular to velocity, doing no work on the particle. Consequently, the speed remains constant while the particle moves in a circular path.
Question 123:
easyA square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $vec{F}$, the net force on the remaining three arms of the loop is:
(2010 Pre)
The net magnetic force on a closed current loop in a uniform magnetic field is always zero.
Therefore, $\vec{F} + \vec{F}_{\text{remaining}} = 0$, which gives $\vec{F}_{\text{remaining}} = -\vec{F}$.
Question 124:
easyTwo long parallel wires are at a distance of $1\text{ m}$. If both of them carry one ampere of current in same direction, then the force of attraction on unit length of the wires will be:
(1998)
The force per unit length between two parallel current-carrying wires is $\frac{F}{l} = \frac{mu_0 I_1 I_2}{2\pi d}$.
Substituting $I_1 = 1\text{ A}$, $I_2 = 1\text{ A}$, and $d = 1\text{ m}$ results in $2 \times 10^{-7}\text{ N/m}$.
Question 125:
easyA beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength $20\text{ Vm}^{-1}$ and $0.5\text{ T}$ at right angles to the direction of motion of the electrons. What is the velocity of the electrons?
(1996)
For undeflected particle motion in perpendicular fields, magnetic force balances electric force ($qE = qvB$).
Thus, velocity $v = \frac{E}{B} = \frac{20}{0.5} = 40\text{ ms}^{-1}$.
Question 126:
easyA current loop in a magnetic field :
(2013)
A current loop in a uniform magnetic field experiences zero net force and can be in equilibrium when its magnetic moment is parallel (stable equilibrium) or antiparallel (unstable equilibrium) to the field.
Question 127:
easyA circular loop of area $0.01 \text{ m}^2$ carrying a current of $10 text{ A}$, is held perpendicular to a magnetic field of intensity $0.1 \text{ T}$. The torque acting on the loop is
(1994)
The torque on a current loop is given by $\tau = M B sin \theta$. Since the loop is held perpendicular to the magnetic field, the area vector is parallel to the field, making $\theta = 0^\circ$ and torque zero.
Question 128:
easyA coil carrying electric current is placed in uniform magnetic field:
(1993)
A current-carrying coil placed in a uniform magnetic field experiences a magnetic torque. No e.m.f. is induced as the magnetic flux linked with the stationary coil remains constant.
Question 129:
easyA current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes :
(1988)
In stable equilibrium, the magnetic moment vector aligns parallel to the magnetic field direction. Consequently, the plane of the coil becomes perpendicular to the magnetic field.
Question 130:
easyA uniform conducting wire of length $12\text{ a}$ and resistance $R$ is wound up as a current carrying coil in the shape of, i. an equilateral triangle of side $a$. ii. a square of side $a$. The magnetic dipole moments of the coil in each case respectively are:
(2021)
For triangle, $N_1 = 4$, $A_1 = \frac{\sqrt{3}}{4}a^2$, so $M_1 = \sqrt{3}Ia^2$. For square, $N_2 = 3$, $A_2 = a^2$, so $M_2 = 3Ia^2$.