JEE Physics Moving Charges and Magnetism MCQs Set 24

Answer: (a) 1 : 4

Explanation:
1. One loop: \( R_1=\frac{L}{2\pi} \), \( B_1=\frac{\mu_0 I}{2R_1}=\frac{\mu_0 I\pi}{L} \).
2. Two loops: \( R_2=\frac{L}{4\pi} \), \( B_2=\frac{2\mu_0 I}{2R_2}=\frac{4\pi\mu_0 I}{L} \).
3. \( B_1:B_2=1:4 \).

Teacher's Note:
1. For a fixed length of wire, B ∝ N² because both N rises and the radius falls as 1/N.
2. Same trick: three turns would give nine times the field.

Answer: (a) the same

Explanation:
1. \( B=\frac{\mu_0 N I}{2R}\propto NI \).
2. New value \( \propto \frac N2\times2I=NI \), unchanged.

Teacher's Note:
1. Only the product N × I matters at the centre for fixed radius.
2. Check what else (radius) is held fixed.

Answer: (c) \( 5\times10^{-6} \) T southward

Explanation:
1. \( B=\frac{\mu_0I}{2\pi r}=\frac{2\times10^{-7}\times100}{4}=5\times10^{-6} \) T.
2. Take east = x, north = y, up = z. Current along \( -\hat x \), point is at \( -\hat z \) from the wire: \( \hat I\times\hat r=(-\hat x)\times(-\hat z)=\hat x\times\hat z=-\hat y \).
3. So the field points south.

Teacher's Note:
1. Use the right-hand thumb rule: thumb along the current, fingers curl in the field direction.
2. Below a west-flowing wire the field points south.

Answer: (d) \( 10^{-4} \)

Explanation:
1. \( B=\frac{\mu_0NI}{2\pi r}=\frac{2\times10^{-7}\times500\times0.1}{0.1} \).
2. \( =10^{-4} \) T.

Teacher's Note:
1. Use n = N/2πr for a toroid.
2. Convert 10 cm to 0.1 m.

Answer: (d) Li++

Explanation:
1. \( F=qvB \) with the same v and B, so \( F\propto q \).
2. Charges: electron e, proton e, He+ e, Li++ 2e. The largest charge is Li++.

Teacher's Note:
1. Force depends on charge only when v and B are equal.
2. Count the number of elementary charges.

Answer: (a) remains constant

Explanation:
1. The magnetic force is always perpendicular to the velocity, so it does no work.
2. Hence speed, and therefore kinetic energy, stays constant.

Teacher's Note:
1. Magnetic force can change direction, never speed.
2. Work done by the force is zero.

Answer: (d) 25 cm

Explanation:
1. \( v=\sqrt{2K/m}=\sqrt{\frac{2\times7.2\times10^{-18}}{9.1\times10^{-31}}}\approx3.98\times10^{6} \) m/s.
2. \( r=\frac{mv}{eB}=\frac{9.1\times10^{-31}\times3.98\times10^{6}}{1.6\times10^{-19}\times9\times10^{-5}}\approx0.25 \) m = 25 cm.

Teacher's Note:
1. Find the speed from the kinetic energy first.
2. Convert metres to centimetres at the end.

Answer: (b) \( q\vec E+q(\vec v\times\vec B) \)

Explanation:
1. The total (Lorentz) force is the electric force \( q\vec E \) plus the magnetic force \( q\vec v\times\vec B \).

Teacher's Note:
1. Order matters in the cross product: v × B, not B × v.
2. This is the full Lorentz force law.

Answer: (c) a straight line

Explanation:
1. At rest the magnetic force is zero, so the charge starts moving along E.
2. Its velocity stays parallel to B, so the magnetic force remains zero and the path is a straight line.

Teacher's Note:
1. Magnetic force needs a velocity component perpendicular to B.
2. Starting from rest, no such component ever appears.

Answer: (c) oscillates

Explanation:
1. Adjacent turns carry current in the same direction and attract, so the spring contracts for steady current.
2. For alternating current the force varies in magnitude (always attractive), so the spring contracts and relaxes periodically, i.e. oscillates.

Teacher's Note:
1. The attractive force depends on I², so it never reverses.
2. For direct current there is a steady contraction.

Answer: (a) repel each other

Explanation:
1. Like charges repel electrostatically.
2. The magnetic attraction between parallel moving charges is smaller by a factor of about v²/c², so the net force is repulsive.

Teacher's Note:
1. In the lab frame the electric repulsion dominates at speeds well below c.
2. Do not stop at the magnetic effect alone.

Answer: (c) \( 1.1\times10^{-3} \) A

Explanation:
1. \( I=ef=1.6\times10^{-19}\times6.8\times10^{15} \).
2. \( I\approx1.09\times10^{-3} \) A.

Teacher's Note:
1. Equivalent current = charge × revolutions per second.
2. The radius is not needed for the current.

Answer: (c) zero

Explanation:
1. Force on each element is directed radially (outwards or inwards).
2. For a closed loop in a uniform field the vector sum of forces is zero (net force = I(∮dl)×B = 0).

Teacher's Note:
1. A closed loop has ∮dl = 0.
2. Torque, not force, is what a uniform field produces on a loop.

Answer: (a) 0

Explanation:
1. \( \tau=MB\sin\theta \), where \( \theta \) is the angle between the moment (normal to the plane) and B.
2. The normal is parallel to B so \( \theta=0 \) and the torque is zero.

Teacher's Note:
1. Torque is maximum when the plane is parallel to the field and zero when it is perpendicular.
2. A trap: do not just multiply NIAB.

Answer: (a) \( \mu\propto n \)

Explanation:
1. \( \mu=\frac{eL}{2m} \) and \( L=\frac{nh}{2\pi} \).
2. So \( \mu\propto n \).

Teacher's Note:
1. The orbital moment is proportional to the angular momentum.
2. Bohr's quantisation gives L ∝ n.

Answer: (c) exerts a force on a charged particle only if it moves across the field lines

Explanation:
1. \( \vec F=q\vec v\times\vec B \) is non-zero only if v has a component perpendicular to B.
2. Motion along the lines gives zero force.

Teacher's Note:
1. No force on a static charge in a magnetic field.
2. Force is perpendicular to both v and B.

Answer: (c) the angle between v and B can have any value other than 0° and 180°

Explanation:
1. \( F=qvB\sin\theta \) is non-zero only if \( \sin\theta\neq0 \).
2. So θ can be anything except 0° and 180°; it need not be 90°.

Teacher's Note:
1. 90° gives the maximum force, not the only possibility.
2. θ = 0° or 180° gives zero force.

Answer: (b) \( \frac{qvR}{2} \)

Explanation:
1. Equivalent current \( I=\frac{q}{T}=\frac{qv}{2\pi R} \).
2. \( \mu=I\pi R^2=\frac{qvR}{2} \).

Teacher's Note:
1. Replace the moving charge by a current loop.
2. Check units: C (m/s) m = A m².

Answer: (c) \( \frac1{R^2} \)

Explanation:
1. \( R=\frac1B\sqrt{\frac{2mV}{q}} \Rightarrow \frac qm=\frac{2V}{B^2R^2} \).
2. So q/m ∝ 1/R².

Teacher's Note:
1. Square the radius expression before inverting.
2. Heavier ions land farther away.

Answer: (c) \( -\vec F \)

Explanation:
1. The net force on a closed loop in a uniform field is zero.
2. So the other three arms together give \( -\vec F \).

Teacher's Note:
1. Total = 0, so the rest = −(force on one arm).
2. Do not multiply by 3.

Answer: (d) \( \frac{\mu_0i}{2\sqrt2R} \)

Explanation:
1. Each semicircle gives \( \frac{\mu_0i}{4R} \), directed along the normal of its own plane (z and y axes respectively).
2. These are perpendicular: \( B=\sqrt2\cdot\frac{\mu_0i}{4R}=\frac{\mu_0i}{2\sqrt2R} \).

Teacher's Note:
1. Add the two perpendicular fields as vectors.
2. A full loop gives μ0 i/2R, a semicircle half of it.

Answer: (b) \( -\vec F \)

Explanation:
1. Arm AB is parallel to B, so the force on it is zero.
2. Net force on the closed loop is zero, so \( \vec F_{AC}=-\vec F_{BC}=-\vec F \).

Teacher's Note:
1. Check which arms are parallel to the field.
2. The magnitudes agree because AC's effective perpendicular length equals BC.