Exams › NEET › Physics › Moving Charges and Magnetism
94 questions with worked solutions.
Answer: B = e/2mπν and K = 2mπ²ν²R²
In a cyclotron, the magnetic field (B) is related to the frequency (ν) by B = e/2mπν, and the kinetic energy (K) of the accelerated particle is given by K = 2mπ²ν²R². These relations are derived from the principles of circular motion and energy in a cyclotron.
Answer: 2 MeV
The radius of a charged particle in a magnetic field depends on its momentum and charge. Since the α-particle has twice the charge and four times the mass of a proton, its energy must be 2 MeV to maintain the same radius.
Answer: move in a circular orbit with its speed unchanged
When a charged particle moves in a magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field (Lorentz force). This force causes the particle to move in a circular path with constant speed, as the magnetic force does no work on the particle.
Answer: it can go undeflected also
An electron can move undeflected if the electric force and magnetic force acting on it are equal and opposite, which occurs when the velocity of the electron satisfies the condition \( v = \frac{E}{B} \).
Answer: 12 cm
The radius of the circular motion of a charged particle in a magnetic field is given by the formula \( r = \frac{mv}{qB} \). Using the kinetic energy \( KE = \frac{1}{2}mv^2 \), we can find \( v = \sqrt{2KE/m} \). Substituting \( KE = 10 \, \text{eV} = 1.6 \times 10^{-18} \, \text{J} \), \( m = 9.1 \times 10^{-31} \, \text{kg} \), \( q = 1.6 \times 10^{-19} \, \text{C} \), and \( B = 10^{-4} \, \text{T} \), the radius calculates to approximately 12 cm.
Answer: 4.0 cm
The radius of the circular path in a magnetic field is directly proportional to the speed of the charged particle. Doubling the speed of the electron will double the radius, making it 4.0 cm.
Answer: 100 keV
The radius of a charged particle in a magnetic field is given by \( r = \frac{mv}{qB} \). For the same radius and magnetic field, \( \frac{mv}{q} \) must be the same. Since kinetic energy \( KE = \frac{1}{2}mv^2 \), we find that \( KE \propto \frac{1}{m} \). The mass of a deuteron is twice that of a proton, so the proton's kinetic energy will be twice that of the deuteron: \( 2 \times 50 \, \text{keV} = 100 \, \text{keV} \).
Answer: 40 m/s
For the electrons to move with constant velocity, the electric force and magnetic force must balance each other. The condition is given by: \( qE = qvB \), where \( v \) is the velocity. Solving for \( v \), we get \( v = \frac{E}{B} = \frac{20}{0.5} = 40 \, \text{m/s} \).
Answer: remain unaffected
The magnetic force on a moving charge is given by the Lorentz force, which depends on the cross product of velocity and magnetic field vectors. Since the velocity and magnetic field are parallel (both along the X-axis), the cross product is zero, and the charge remains unaffected.
Answer: B̅ should be perpendicular to the direction of velocity and E̅ should be along the direction of velocity.
For the particle to move in a horizontal direction without deviation, the magnetic force (qvB) and electric force (qE) must cancel each other. This requires the magnetic field to be perpendicular to the velocity and the electric field to be along the velocity, ensuring no net force acts perpendicular to the motion.
Answer: μ₀i / 2√2R
Each semicircular loop produces a magnetic field of magnitude μ₀i / 4R at the center, but the directions are perpendicular to each other due to their orientations in the x-y and x-z planes. The resultant field is the vector sum, which gives μ₀i / 2√2R.
Answer: perpendicular to the magnetic field
A current-carrying coil in a uniform magnetic field experiences a torque that tends to align its plane perpendicular to the magnetic field to minimize potential energy.
Answer: angle between v̅ and B̅ can have any value other than zero and 180°
The magnetic force on a charged particle is given by F̅ = q(v̅ × B̅), which depends on the sine of the angle between v̅ and B̅. The force is non-zero when the angle is neither 0° nor 180°, as sin(0°) = sin(180°) = 0.
Answer: putting in parallel a resistance of 15Ω
To convert the galvanometer into an ammeter, a shunt resistance is connected in parallel. The shunt resistance is calculated using the formula Rs = (Ig * Rg) / (I - Ig), where Ig is the galvanometer current, Rg is the galvanometer resistance, and I is the desired current. Substituting the values, Rs = (1 * 60) / (5 - 1) = 15Ω. Thus, a 15Ω resistance in parallel is required.
Answer: 2450 Ω in series
The full-scale deflection current is 25 × 4 × 10⁻⁴ A = 0.01 A. To convert the galvanometer into a voltmeter with a range of 25 V, the series resistance required is R = (V/I) - G = (25/0.01) - 50 = 2450 Ω.
Answer: 0.5 amp
The current through the shunt is calculated using the current division rule. Since the shunt resistance is 2 ohms and the galvanometer resistance is 8 ohms, the current through the shunt is (8 / (8 + 2)) × 1 = 0.8 × 1 = 0.8 A. Thus, the correct answer is 0.8 A.
Q17. A galvanometer acting as a voltmeter will have
Answer: a high resistance in series with its coil
To convert a galvanometer into a voltmeter, a high resistance is connected in series with its coil to limit the current passing through it, allowing it to measure potential difference accurately.
Answer: r = (4π × 10⁻⁷ × 12) / (2π × 3 × 10⁻⁵)
The magnetic field due to a long straight current-carrying conductor is given by B = μ₀I / 2πr. Rearranging for r, we get r = μ₀I / 2πB. Substituting μ₀ = 4π × 10⁻⁷, I = 12 A, and B = 3 × 10⁻⁵ Wb/m², the correct expression matches option B.
Answer: τ = 2000 × 2 × 1.5 × 10⁻⁴ × 5 × 10⁻² × 1/2
The torque on a solenoid is given by τ = MB sin θ, where M = nIA. Substituting the given values (n = 2000, I = 2 A, A = 1.5 × 10⁻⁴ m², B = 5 × 10⁻² T, and sin 30° = 1/2), we get τ = 2000 × 2 × 1.5 × 10⁻⁴ × 5 × 10⁻² × 1/2.
Answer: True
To convert a galvanometer into a voltmeter, a high resistance is connected in series with it. This limits the current passing through the galvanometer, allowing it to measure potential difference accurately.