MDCAT Physics chapter 10: Physics · Electromagnetism MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
18 exam-style questions on Physics · Electromagnetism, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Ten Physics · Electromagnetism questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. In a moving-coil galvanometer the pole pieces are concave and a soft-iron cylinder sits inside the coil, so the magnetic field is radial. The main purpose of this is to
A
make the deflection proportional to the current
B
cancel the field due to the current in the coil
C
reduce the resistance of the galvanometer coil
D
stop the coil from oscillating about its position
Show answer
Answer: A. In a radial field the plane of the coil always lies along the field, so the deflecting torque BINA does not change as the coil turns. The deflection θ is then proportional to I, giving an evenly spaced scale.
2. For a moving-coil galvanometer, I = (c/BAN)θ, where c is the couple per unit twist of the suspension. Its sensitivity (deflection per unit current) can be increased by using
A
a weaker magnet, so that the coil turns more freely
B
a coil with fewer turns, so that it is lighter
C
a coil of smaller area, so that it turns easily
D
a suspension with a smaller value of c
Show answer
Answer: D. θ/I = BAN/c, so the sensitivity rises if B, A or N is made larger or c is made smaller. A weaker field, fewer turns or a smaller coil all reduce the deflecting torque for a given current.
3. The magnetic field lines around a long straight wire carrying a steady current are
A
concentric circles round the wire
B
straight lines parallel to the wire
C
straight lines radiating out from the wire
D
a single helix coiled along the wire
Show answer
Answer: A. The field of a straight current forms concentric circles in planes perpendicular to the wire, with the direction given by the right-hand grip rule. The field is weaker further from the wire (B ∝ 1/r).
4. A solenoid 0.20 m long has 400 turns and carries a current of 5.0 A. Taking μ0 = 4π × 10−7 T m A−1, the magnetic field inside it is
A
8π × 10−4 T
B
4π × 10−3 T
C
2π × 10−3 T
D
8π × 10−3 T
Show answer
Answer: B. n = 400/0.20 = 2000 turns per metre, so B = μ0nI = 4π × 10−7 × 2000 × 5.0 = 4π × 10−3 T (about 0.013 T). Using N = 400 in place of n gives 8π × 10−4 T.
5. In an experiment to find e/m, electrons accelerated from rest through a potential difference V move in a circle of radius r at right angles to a uniform magnetic field B. The value of e/m is
A
V/(2B2r2)
B
B2r2/(2V)
C
2V/(B2r2)
D
2V/(Br)
Show answer
Answer: C. eV = ½mv2 and r = mv/eB, so v = eBr/m. Substituting gives eV = e2B2r2/2m, so e/m = 2V/(B2r2). B2r2/(2V) is m/e, the reciprocal.
6. A horizontal wire carries a current from west to east in a uniform magnetic field directed vertically downwards. The magnetic force on the wire is directed towards the
A
north
B
south
C
east
D
vertically upwards
Show answer
Answer: A. F = I(L × B): with L pointing east and B pointing down, east × down points north (Fleming's left-hand rule gives the same). The force is perpendicular to both the current and the field, so it cannot be vertical or along the wire.
7. A horizontal straight wire of mass 20 g and length 0.50 m lies at right angles to a horizontal uniform magnetic field of 0.49 T. Taking g = 9.8 m s−2, the current that makes the magnetic force on the wire just balance its weight is
A
0.80 A
B
0.40 A
C
8.0 A
D
0.082 A
Show answer
Answer: A. For balance BIL = mg, so I = (0.020 × 9.8) ÷ (0.49 × 0.50) = 0.196 ÷ 0.245 = 0.80 A. Leaving out L gives 0.40 A, and forgetting g gives 0.082 A.
8. A straight wire carrying a current lies in a uniform magnetic field. The force on the wire is greatest when the wire is
A
parallel to the field
B
perpendicular to the field
C
at 45° to the field
D
antiparallel to the field
Show answer
Answer: B. F = ILB sin θ is greatest when θ = 90°. When the wire is parallel or antiparallel to the field (θ = 0° or 180°) the force is zero.
9. Two long parallel wires 0.20 m apart in air each carry a current of 10 A. Taking μ0 = 4π × 10−7 T m A−1, the force on each metre length of either wire is
A
1.0 × 10−5 N
B
2.0 × 10−4 N
C
1.0 × 10−4 N
D
6.3 × 10−4 N
Show answer
Answer: C. The field of one wire at the other is B = μ0I/2πd = 2 × 10−7 × 10 ÷ 0.20 = 1.0 × 10−5 T, and the force on 1 m of the other wire is BIL = 1.0 × 10−5 × 10 × 1 = 1.0 × 10−4 N. 1.0 × 10−5 N forgets the second current, and leaving out the 2π gives 6.3 × 10−4 N.
10. A proton and an alpha particle enter the same uniform magnetic field at right angles to it with equal kinetic energies. Take the mass of the alpha particle as 4 times, and its charge as 2 times, those of the proton. The ratio of the radius of the alpha particle's path to that of the proton's path is
A
2 : 1
B
1 : 2
C
1 : 1
D
4 : 1
Show answer
Answer: C. r = mv/qB = √(2mK)/qB. For the alpha particle this is √(2 × 4m × K)/(2eB) = √(2mK)/(eB), exactly the proton's value, so the ratio is 1 : 1. 2 : 1 would be right for equal speeds, not equal energies.
Physics · Electromagnetism: questions students ask
How many Physics · Electromagnetism MCQs are there?
18, each with the answer and a short explanation, across 2 topics: Magnetic fields and forces, Charges in magnetic fields. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Physics chapter 10 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
