Moving Charges and Magnetism
Class 12 Physics | JEE Advanced Level
Q1.
A proton moving with speed 2 × 106 m s−1 enters a uniform magnetic field of 0.5 T at an angle 30° with the field. The radius of the helical path is closest to:
Answer: A
Only perpendicular velocity causes circular motion. v⊥ = v sin30° = 1 × 106 m s−1.
r = mv⊥ / qB = (1.67 × 10−27 × 106)/(1.6 × 10−19 × 0.5) ≈ 2.09 × 10−2 m = 2.1 cm.
Only perpendicular velocity causes circular motion. v⊥ = v sin30° = 1 × 106 m s−1.
r = mv⊥ / qB = (1.67 × 10−27 × 106)/(1.6 × 10−19 × 0.5) ≈ 2.09 × 10−2 m = 2.1 cm.
Q2.
An electron and a proton enter the same uniform magnetic field with equal kinetic energies and perpendicular velocities. The ratio of their circular radii re : rp is:
Answer: B
For same K and same |q|, r = p/qB = √(2mK)/qB. Hence r ∝ √m. Therefore re/rp = √(me/mp) = 1/√1836.
For same K and same |q|, r = p/qB = √(2mK)/qB. Hence r ∝ √m. Therefore re/rp = √(me/mp) = 1/√1836.
Q3.
A charged particle moves in a uniform magnetic field with pitch p and radius r. If its speed is doubled keeping the angle with magnetic field unchanged, the new pitch and radius are:
Answer: B
r ∝ v⊥ and pitch = v∥T. Since T = 2πm/qB is unchanged and both velocity components double, radius and pitch both double.
r ∝ v⊥ and pitch = v∥T. Since T = 2πm/qB is unchanged and both velocity components double, radius and pitch both double.
Q4.
A particle of charge q and mass m is accelerated through potential V and then enters a magnetic field B normally. Its radius is:
Answer: B
qV = ½mv² ⇒ v = √(2qV/m). Radius r = mv/qB = (1/B)√(2mV/q).
qV = ½mv² ⇒ v = √(2qV/m). Radius r = mv/qB = (1/B)√(2mV/q).
Q5.
In a velocity selector, electric field E and magnetic field B are perpendicular. Ions pass undeviated. If both E and B are doubled, selected speed becomes:
Answer: B
For no deflection, qE = qvB ⇒ v = E/B. Doubling both E and B keeps E/B unchanged.
For no deflection, qE = qvB ⇒ v = E/B. Doubling both E and B keeps E/B unchanged.
Q6.
A wire of length l carries current I along +x-axis. Magnetic field is B = B0(ĵ + k̂). Force on wire is:
Answer: B
F = I L × B = I(l î) × B0(ĵ + k̂) = IlB0(k̂ − ĵ).
F = I L × B = I(l î) × B0(ĵ + k̂) = IlB0(k̂ − ĵ).
Q7.
A semicircular wire of radius R carries current I. It is placed in a uniform magnetic field B perpendicular to its plane. The magnitude of force on the curved part is:
Answer: B
Net force on any current-carrying wire in uniform B depends on end-to-end displacement. For semicircle, effective length is diameter 2R. Hence F = I(2R)B = 2IRB.
Net force on any current-carrying wire in uniform B depends on end-to-end displacement. For semicircle, effective length is diameter 2R. Hence F = I(2R)B = 2IRB.
Q8.
A circular loop of radius R carries current I. It is placed in a uniform magnetic field B with its magnetic moment making 60° with B. Torque is:
Answer: B
τ = MB sinθ = IA B sin60° = IπR²B × √3/2.
τ = MB sinθ = IA B sin60° = IπR²B × √3/2.
Q9.
A moving coil galvanometer has current sensitivity Si. If number of turns and area are both doubled while torsional constant remains same, new current sensitivity is:
Answer: C
Current sensitivity θ/I = NBA/k. If N and A are both doubled, sensitivity becomes 4 times.
Current sensitivity θ/I = NBA/k. If N and A are both doubled, sensitivity becomes 4 times.
Q10.
A galvanometer of resistance 20 Ω gives full scale deflection at 2 mA. The shunt required to convert it into an ammeter of range 1 A is approximately:
Answer: B
Voltage across galvanometer at full scale = IgG = 0.002 × 20 = 0.04 V. Shunt current = 1 − 0.002 = 0.998 A. S = 0.04/0.998 ≈ 0.040 Ω.
Voltage across galvanometer at full scale = IgG = 0.002 × 20 = 0.04 V. Shunt current = 1 − 0.002 = 0.998 A. S = 0.04/0.998 ≈ 0.040 Ω.
Q11.
Two long parallel wires carry currents 3 A and 5 A in opposite directions. Distance between them is 20 cm. Force per unit length is:
Answer: B
Opposite currents repel. F/L = μ0I1I2/(2πd) = (2 × 10−7 × 15)/0.2 = 1.5 × 10−5 N m−1.
Opposite currents repel. F/L = μ0I1I2/(2πd) = (2 × 10−7 × 15)/0.2 = 1.5 × 10−5 N m−1.
Q12.
Three long parallel wires are placed at the vertices of an equilateral triangle of side a. Equal currents I flow out of the page in all wires. The magnetic force per unit length on one wire has magnitude:
Answer: B
Each of two forces has magnitude f = μ0I²/(2πa), attractive along sides, angle 60° between them. Resultant = √(f² + f² + 2f²cos60°) = √3 f.
Each of two forces has magnitude f = μ0I²/(2πa), attractive along sides, angle 60° between them. Resultant = √(f² + f² + 2f²cos60°) = √3 f.
Q13.
Magnetic field at the centre of a circular loop of radius R carrying current I is B. The field at the centre if the loop is reshaped into a square using the same wire and current is:
Answer: C
Same wire length: 2πR = 4a ⇒ side of square a = πR/2. Field at centre of square = 4 × [μ0I/(4πd)](sin45° + sin45°), where d = a/2. Thus Bs = 2√2 μ0I/(πa) = 4√2 μ0I/(π²R). Since circular loop field B = μ0I/(2R), Bs/B = 8√2/π².
Same wire length: 2πR = 4a ⇒ side of square a = πR/2. Field at centre of square = 4 × [μ0I/(4πd)](sin45° + sin45°), where d = a/2. Thus Bs = 2√2 μ0I/(πa) = 4√2 μ0I/(π²R). Since circular loop field B = μ0I/(2R), Bs/B = 8√2/π².
Q14.
A long straight wire carries current I. A point P is at distance a from it. The magnetic field at P is B. If current is doubled and distance is tripled, field becomes:
Answer: A
For long wire, B ∝ I/r. New B = B × (2/3) = 2B/3.
For long wire, B ∝ I/r. New B = B × (2/3) = 2B/3.
Q15.
A current I flows through a circular arc of radius R subtending angle 120° at the centre. Magnetic field at centre due to the arc is:
Answer: A
For arc angle θ in radians, B = μ0Iθ/(4πR). Here θ = 120° = 2π/3. Hence B = μ0I/(6R).
For arc angle θ in radians, B = μ0Iθ/(4πR). Here θ = 120° = 2π/3. Hence B = μ0I/(6R).
Q16.
Two concentric circular coils of radii R and 2R carry currents I and 2I in opposite directions. Magnetic field at common centre is:
Answer: A
Field due to smaller coil = μ0I/(2R). Field due to larger coil = μ0(2I)/(2×2R) = μ0I/(2R). Opposite directions, so net field is zero.
Field due to smaller coil = μ0I/(2R). Field due to larger coil = μ0(2I)/(2×2R) = μ0I/(2R). Opposite directions, so net field is zero.
Q17.
A solenoid has n turns per unit length and carries current I. A particle of charge q moves along the solenoid axis with speed v. Magnetic force on particle is:
Answer: C
Inside a long solenoid, B is along its axis. If particle velocity is also along the axis, angle between v and B is zero. Magnetic force qvB sin0° = 0.
Inside a long solenoid, B is along its axis. If particle velocity is also along the axis, angle between v and B is zero. Magnetic force qvB sin0° = 0.
Q18.
A charged particle enters a region where E and B are parallel. If initial velocity is perpendicular to both fields, the particle trajectory is generally:
Answer: C
Magnetic field gives circular motion perpendicular to B. Electric field parallel to B accelerates particle along B, so parallel velocity changes with time. Hence helix with changing pitch.
Magnetic field gives circular motion perpendicular to B. Electric field parallel to B accelerates particle along B, so parallel velocity changes with time. Hence helix with changing pitch.
Q19.
A rectangular loop of sides a and b carries current I in uniform magnetic field B. The plane of loop is parallel to B. Torque is:
Answer: A
Magnetic moment M = IA normal to plane. If plane is parallel to B, normal is perpendicular to B. Thus τ = MB sin90° = IabB.
Magnetic moment M = IA normal to plane. If plane is parallel to B, normal is perpendicular to B. Thus τ = MB sin90° = IabB.
Q20.
A magnetic dipole of moment M is rotated from stable equilibrium to unstable equilibrium in a uniform magnetic field B. Work done is:
Answer: B
Potential energy U = −MB cosθ. Stable: θ = 0 ⇒ U = −MB. Unstable: θ = π ⇒ U = +MB. Work done = ΔU = 2MB.
Potential energy U = −MB cosθ. Stable: θ = 0 ⇒ U = −MB. Unstable: θ = π ⇒ U = +MB. Work done = ΔU = 2MB.
Q21.
A particle moves in a circular path under magnetic field. If its kinetic energy becomes four times, radius becomes:
Answer: C
r = √(2mK)/(qB), so r ∝ √K. If K becomes 4K, radius becomes 2r.
r = √(2mK)/(qB), so r ∝ √K. If K becomes 4K, radius becomes 2r.
Q22.
A charged particle of mass m and charge q enters a magnetic field B normally. Time taken to complete half circle is:
Answer: A
Cyclotron period T = 2πm/qB. Half circle time = T/2 = πm/qB.
Cyclotron period T = 2πm/qB. Half circle time = T/2 = πm/qB.
Q23.
In a cyclotron, if magnetic field is doubled while oscillator frequency is unchanged, resonance for same particle:
Answer: B
Cyclotron frequency f = qB/(2πm). If B doubles, required oscillator frequency doubles. If unchanged, resonance condition fails.
Cyclotron frequency f = qB/(2πm). If B doubles, required oscillator frequency doubles. If unchanged, resonance condition fails.
Q24.
A wire carrying current I is bent into a regular hexagon of side a. Magnetic field at centre is:
Answer: B
For each side, perpendicular distance from the centre is d = a√3/2 and the end angles are 30° and 30°. Field due to one side is μ0I/(4πd)(sin30° + sin30°) = μ0I/(4πd). For six sides, B = 6μ0I/(4πd) = 3μ0I/(2πd) = √3 μ0I/(πa).
For each side, perpendicular distance from the centre is d = a√3/2 and the end angles are 30° and 30°. Field due to one side is μ0I/(4πd)(sin30° + sin30°) = μ0I/(4πd). For six sides, B = 6μ0I/(4πd) = 3μ0I/(2πd) = √3 μ0I/(πa).
Q25.
A straight conductor of length 0.5 m carries current 4 A. It makes angle 30° with a magnetic field of 0.2 T. Magnetic force is:
Answer: B
F = BIL sinθ = 0.2 × 4 × 0.5 × sin30° = 0.2 N.
F = BIL sinθ = 0.2 × 4 × 0.5 × sin30° = 0.2 N.
Q26.
A current loop is in uniform magnetic field. Which statement is correct?
Answer: A
In uniform magnetic field, force on opposite elements cancels, so net force on closed loop is zero. Torque τ = M × B may be non-zero.
In uniform magnetic field, force on opposite elements cancels, so net force on closed loop is zero. Torque τ = M × B may be non-zero.
Q27.
A loop of area A and current I is placed in magnetic field B. If magnetic potential energy is minimum, angle between magnetic moment and B is:
Answer: A
U = −MB cosθ. Minimum value is −MB, obtained when cosθ = 1, so θ = 0°.
U = −MB cosθ. Minimum value is −MB, obtained when cosθ = 1, so θ = 0°.
Q28.
A charged particle moving with velocity v = 3î + 4ĵ enters magnetic field B = 2k̂. If charge q = 5 C, force vector is:
Answer: A
F = q(v × B). (3î + 4ĵ) × 2k̂ = 8î − 6ĵ. Multiplying by q = 5 gives F = 40î − 30ĵ.
F = q(v × B). (3î + 4ĵ) × 2k̂ = 8î − 6ĵ. Multiplying by q = 5 gives F = 40î − 30ĵ.
Q29.
A particle has velocity vector exactly opposite to uniform magnetic field. The path of the particle is:
Answer: C
Magnetic force is qvB sinθ. Here θ = 180°, so sinθ = 0. Hence no force and particle moves straight.
Magnetic force is qvB sinθ. Here θ = 180°, so sinθ = 0. Hence no force and particle moves straight.
Q30.
A proton and an alpha particle enter the same magnetic field normally with same speed. Ratio of radii rp : rα is:
Answer: B
r = mv/qB. Alpha particle has mass 4mp and charge 2e. rα = (4mpv)/(2eB)=2rp. Ratio = 1:2.
r = mv/qB. Alpha particle has mass 4mp and charge 2e. rα = (4mpv)/(2eB)=2rp. Ratio = 1:2.
Q31.
An electron enters a magnetic field normally. Its angular frequency is:
Answer: A
Angular cyclotron frequency ω = qB/m. For electron magnitude is eB/m.
Angular cyclotron frequency ω = qB/m. For electron magnitude is eB/m.
Q32.
If a charged particle moves undeflected through crossed E and B fields, then after entering only magnetic field B, radius of path is:
Answer: A
Velocity selector gives v = E/B. In magnetic field alone, r = mv/qB = mE/(qB²).
Velocity selector gives v = E/B. In magnetic field alone, r = mv/qB = mE/(qB²).
Q33.
A circular coil has N turns, radius R and carries current I. Magnetic field at centre is B. If radius is doubled and number of turns is halved, current unchanged, new field is:
Answer: A
B = μ0NI/(2R). New B′ = μ0(N/2)I/[2(2R)] = B/4.
B = μ0NI/(2R). New B′ = μ0(N/2)I/[2(2R)] = B/4.
Q34.
The magnetic field on the axis of a circular loop at distance x from centre is maximum at:
Answer: A
B = μ0IR²/[2(R²+x²)3/2]. Denominator is minimum at x = 0, hence field is maximum at centre.
B = μ0IR²/[2(R²+x²)3/2]. Denominator is minimum at x = 0, hence field is maximum at centre.
Q35.
A wire carrying current I is bent into a circle. If the same wire is bent into a circle of half radius by using only half of the wire length and same current, field at centre becomes:
Answer: C
For a circular loop, B = μ0I/(2R). If radius becomes R/2 with same current, field becomes 2B.
For a circular loop, B = μ0I/(2R). If radius becomes R/2 with same current, field becomes 2B.
Q36.
A charged particle completes 10 revolutions in 2 μs in a uniform magnetic field. Its cyclotron frequency is:
Answer: B
Frequency = number of revolutions/time = 10/(2 × 10−6) = 5 × 106 Hz = 5 MHz.
Frequency = number of revolutions/time = 10/(2 × 10−6) = 5 × 106 Hz = 5 MHz.
Q37.
A current carrying loop has magnetic moment M. It is placed in field B. If it is rotated through small angle θ from stable equilibrium, restoring torque is approximately:
Answer: B
Torque magnitude τ = MB sinθ ≈ MBθ. Restoring torque acts opposite to displacement, so τ = −MBθ.
Torque magnitude τ = MB sinθ ≈ MBθ. Restoring torque acts opposite to displacement, so τ = −MBθ.
Q38.
A galvanometer has resistance G and full-scale current Ig. To convert into voltmeter of range V, series resistance should be:
Answer: A
At full scale, V = Ig(G + R). Therefore R = V/Ig − G.
At full scale, V = Ig(G + R). Therefore R = V/Ig − G.
Q39.
A galvanometer has resistance 50 Ω and full scale current 1 mA. Series resistance required for 10 V voltmeter is:
Answer: A
R = V/Ig − G = 10/0.001 − 50 = 10000 − 50 = 9950 Ω.
R = V/Ig − G = 10/0.001 − 50 = 10000 − 50 = 9950 Ω.
Q40.
A particle enters a magnetic field with velocity perpendicular to B. Which quantity remains constant?
Answer: C
Magnetic force is always perpendicular to velocity, so it does no work. Speed and kinetic energy remain constant, but velocity and momentum directions change.
Magnetic force is always perpendicular to velocity, so it does no work. Speed and kinetic energy remain constant, but velocity and momentum directions change.
Q41.
A square loop of side a carries current I. Magnetic field at its centre is:
Answer: A
For one side, d = a/2 and θ = 45°. Bone side = μ0I/(4πd)(2sin45°). Four sides give B = 2√2 μ0I/(πa).
For one side, d = a/2 and θ = 45°. Bone side = μ0I/(4πd)(2sin45°). Four sides give B = 2√2 μ0I/(πa).
Q42.
A current carrying wire is placed along magnetic field. The force on it is:
Answer: B
F = BIL sinθ. If wire is along field, θ = 0°, so F = 0.
F = BIL sinθ. If wire is along field, θ = 0°, so F = 0.
Q43.
A loop has magnetic moment 0.2 A m² and is placed in 5 T field. Maximum torque is:
Answer: B
Maximum torque τmax = MB = 0.2 × 5 = 1 N m.
Maximum torque τmax = MB = 0.2 × 5 = 1 N m.
Q44.
Two ions have same charge but masses m and 4m. They are accelerated by same potential and enter same B normally. Ratio of radii is:
Answer: A
r = (1/B)√(2mV/q). Hence r ∝ √m. For masses m and 4m, ratio = 1:2.
r = (1/B)√(2mV/q). Hence r ∝ √m. For masses m and 4m, ratio = 1:2.
Q45.
A long wire carries current I upward. At a point to the east of wire, direction of magnetic field is:
Answer: A
Use right-hand thumb rule. Thumb upward along current; curled fingers give magnetic field. At east side, field points north.
Use right-hand thumb rule. Thumb upward along current; curled fingers give magnetic field. At east side, field points north.
Q46.
The magnetic field inside a long solenoid is B = μ0nI. If length and number of turns are both doubled, keeping current same, field becomes:
Answer: B
n = N/L. If both N and L are doubled, n remains same. Hence B remains unchanged.
n = N/L. If both N and L are doubled, n remains same. Hence B remains unchanged.
Q47.
A charged particle moves in a magnetic field. Its magnetic force can change:
Answer: C
Magnetic force is perpendicular to velocity, so it changes direction of velocity but not speed or kinetic energy.
Magnetic force is perpendicular to velocity, so it changes direction of velocity but not speed or kinetic energy.
Q48.
A current loop is placed in uniform magnetic field with magnetic moment anti-parallel to B. The equilibrium is:
Answer: B
Anti-parallel position has maximum potential energy U = +MB. Small rotation lowers energy, so it is unstable equilibrium.
Anti-parallel position has maximum potential energy U = +MB. Small rotation lowers energy, so it is unstable equilibrium.
Q49.
Magnetic field due to a very long straight wire at distance r is B. At distance 2r, the magnetic energy density becomes:
Answer: B
B ∝ 1/r. At 2r, field becomes B/2. Magnetic energy density u = B²/(2μ0), so new density = u/4.
B ∝ 1/r. At 2r, field becomes B/2. Magnetic energy density u = B²/(2μ0), so new density = u/4.
Q50.
A beam of positive ions enters a region with magnetic field into the page. If the ions initially move to the right, they are deflected:
Answer: A
For positive charge, F = q(v × B). v is right (+x), B is into page (−z). Hence v × B = +y, i.e. upward.
For positive charge, F = q(v × B). v is right (+x), B is into page (−z). Hence v × B = +y, i.e. upward.