A magnetic field:
The magnetic force is given by \(F = qvB\sin\theta\). If the particle moves across the field lines, \(\sin\theta \neq 0\), resulting in a non-zero force.
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.
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.
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:
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 charge \( q = 1.6 \times 10^{-12} \text{ C} \) moving with speed of \( v \text{ m s}^{-1} \) crosses electric field \( |\vec{E}| = 6 \times 10^4 \text{ V m}^{-1} \) and magnetic field \( |\vec{B}| = 1.2 \text{ T} \). The electric field and magnetic fields are crossed and velocity \( v \) is also perpendicular to both. If the charge particle crosses both fields undeflected, the value of \( v \) is
For a particle to cross perpendicular electric and magnetic fields undeflected, the net force must be zero, which requires \( qE = qvB ⇒ v = \frac{E}{B} \). Substituting the given values: \( v = \frac{6 \times 10^4}{1.2} = 5 \times 10^4 \text{ m/s} \).
Consider the following statements:
A. Magnetic force on a moving charged particle is always non-zero.
B. Magnetic force can change kinetic energy of a charged particle.
C. Magnetic force can change linear momentum of a charged particle.
D. A charged particle at rest does not feel magnetic force on it.
The correct statement(s) is/are
Magnetic force is perpendicular to velocity, so it does no work and KE remains constant. It can change the direction of velocity (and thus momentum). For a particle at rest (\(v = 0\)), magnetic force is zero.
Assertion (A): A magnetic field can accelerate a charge particle.
Reason (R): A steady current carrying wire does not generate an electric field outside it.
In the light of the above statements, choose the correct answer from the options given below.
A magnetic field exerts a force perpendicular to velocity, changing its direction, hence accelerating the particle. Outside a steady current-carrying wire, there is no net charge, so the electric field is zero. Both statements are true, but they are unrelated.
In crossed electric and magnetic field, the velocity of charged particle which passes undeflected through the region may be (where \(E\) is electric field and \(B\) is magnetic field)
For a charged particle to pass undeflected in crossed fields, the electric force must balance the magnetic force: \(qE = qvB ⇒ v = \frac{E}{B}\).
A charged particle enters a magnetic field at right angles to the magnetic field. The field exists for a length equal to 1.5 times the radius of circular path of the circle. The particle will be deviated from its path by angle
Since the width of the magnetic field \(d = 1.5R > R\), the particle cannot cross the field to the other side. It will complete a semi-circular path inside the field and emerge from the same side it entered, yielding a deviation of \(180^\circ\).