Class 12 Physics, Chapter 4 (NCERT). 15 questions by concept, 1 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
A straight wire carries 4 A in a field of 3 T, making 30° with the field. What is the magnetic force per unit length?
(a) 6 N/m
F/L=IB sin30°=IB/2; use the perpendicular current component.
A 10 cm straight wire carries 10 A parallel to a uniform field of 2 T. What is its magnetic force?
(d) Zero N
The angle is zero, so I L×B vanishes even though current, length and field are nonzero.
The diagram shows an electron velocity and a uniform magnetic field in the page plane. The x-axis points right, y-axis points up, and +z is out of the page. In which direction is its magnetic force?
(d) Towards -z
The cross product +x×+y points +z. Multiplication by the negative electron charge reverses it to -z.
A proton moves in a circle under only a uniform magnetic field. After a quarter revolution, which pair of changes is correct?
(a) Velocity changes; kinetic energy stays constant
The force continuously changes direction of velocity but does zero work because it is perpendicular to velocity.
In a fixed uniform B, a nonrelativistic particle enters perpendicularly at speed v and cyclotron frequency f. Its speed is then increased to 3v. What is the new frequency?
(b) f
Frequency |q|B/(2πm) is independent of speed. The radius increases with speed so the ratio v/r stays constant.
Two electrons move nonrelativistically in the same uniform field, perpendicular to B. Their speeds are v and 2v. What is T_fast:T_slow?
(b) 1:1
Period 2πm/(eB) depends on mass, charge and field, not speed.
Two concentric coils have perpendicular axes. At their common centre, each produces magnetic-field magnitude B0. What factor must multiply both currents to make the resultant field magnitude B0 at that centre, with geometry unchanged?
(b) 1/√2
This calculation requires the two fields to be specified at the same point. For concentric coils with perpendicular axes, at the common centre each contributes magnitude along a perpendicular direction.
Their initial resultant is .
With geometry fixed, each field scales linearly with current. Applying the same factor f to both currents gives resultant .
Set , so . The separate centre-fields of nonconcentric coils would not determine this result at an unspecified point.
Two perpendicular coils create fields 6 μT along +x and 8 μT along +y at their common centre. What is the resultant magnitude?
(b) 10 μT
Orthogonal contributions combine through Pythagoras: sqrt(6²+8²)=10 μT. Directly adding their magnitudes ignores their directions.
Within a long wire of radius a carrying uniform current density, move from radius a/4 to a/2. What is B_final:B_initial?
(a) 2:1
Both points are inside. The field increases linearly with radius, so doubling radius doubles B.
A loop is traversed anticlockwise as viewed on the page. Currents 5 A out of the page and 2 A into the page pierce its interior. What is the circulation ∮B·dl?
(d) 3μ0 A
For anticlockwise traversal as seen on the page, the positive surface normal points out of the page by the right-hand rule.
Signed enclosed current is I_enc = 5 − 2 A.
Ampère’s law gives ∮B·dl = μ₀(5 − 2) A, with units T m. Add currents with their signs, not just their magnitudes.
Therefore, the requested answer is 3μ0 A.
A semicircular arc carries current I and has radius R. Only its contribution is considered. What is its magnetic-field magnitude at the centre of curvature?
(b) μ0I/(4R)
A semicircle subtends π radians, half a full loop, so its field is half μ0I/(2R).
A long air-core solenoid has 500 turns over 50 cm and carries 5 A. Use μ0=4π×10^-7 T m/A. What is the approximate central field?
(d) 2000π μT
Use B=μ0(N/L)I with L in metres. Total turns alone do not specify the winding density.
Two long parallel wires carry steady currents in the same direction. What mutual force do they exert?
(a) Attraction
The field of each wire and the current direction in the other give a force toward the other wire.
An electron moves anticlockwise around a circular orbit as viewed from above the page. Which pair of directions is correct?
(a) Angular momentum out of page; magnetic moment into page
Angular momentum follows the electron motion by the right-hand rule. Conventional current is opposite electron motion, so magnetic moment points oppositely.
In a moving-coil galvanometer, a radial magnetic field stays in the plane of the coil. What is the magnetic torque for current I?
(a) NIAB
The magnetic moment is normal to the plane and perpendicular to radial B, giving sin90°=1.
Wrong
A positive charge is pushed in the direction the magnetic field points.
Why it feels right. An electric force on a positive charge points along the electric field, so I reuse that rule.
Right
The magnetic force is perpendicular to velocity and field and vanishes when they are parallel.
Worked example
