Fields Practice Questions
IB (International Baccalaureate) · IB Physics HL · 148 free MCQs with instant results and detailed explanations.
148
Total
51
Easy
72
Medium
25
Hard
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Sample Questions from Fields
Here are 10 sample questions. Start a quiz to get randomized questions with scoring.
Q1
Easy
What is the gravitational field strength at a distance of 10 meters from a mass of 100 kg?
A.
0.1 N/kg
B.
1 N/kg
C.
9.8 N/kg
D.
10 N/kg
Show Answer & Explanation
Correct Answer: A
The gravitational field strength (g) can be calculated using the formula g = G * M / rยฒ. Here, G is the gravitational constant (6.67 x 10^-11 N mยฒ/kgยฒ), M is 100 kg, and r is 10 m. This results in a value of 0.1 N/kg.
Q2
Easy
Which of the following statements correctly describes electric field lines?
A.
They can cross each other.
B.
They are always straight lines.
C.
They point away from positive charges.
D.
They are shorter than magnetic field lines.
Show Answer & Explanation
Correct Answer: C
Electric field lines indicate the direction of the electric field; they emanate from positive charges and terminate on negative charges, hence they point away from positive charges.
Q3
Easy
A proton is placed in a uniform magnetic field. If it moves perpendicular to the field lines, what will be the effect on its motion?
A.
It will continue in a straight line.
B.
It will accelerate in the direction of the field.
C.
It will experience a force perpendicular to its velocity.
D.
It will stop moving.
Show Answer & Explanation
Correct Answer: C
When a charged particle like a proton moves perpendicular to magnetic field lines, it experiences a magnetic force that acts perpendicular to both its velocity and the magnetic field, causing it to move in a circular path.
Q4
Medium
A charged particle moves in a uniform magnetic field of strength 0.5 T. If the particle has a charge of 2 C and moves with a velocity of 10 m/s, what is the magnetic force acting on it?
A.
10 N
B.
5 N
C.
20 N
D.
0 N
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Correct Answer: A
The magnetic force (F) on a charged particle can be calculated using F = qvB sin(ฮธ). Here, ฮธ is 90 degrees since the particle moves perpendicular to the field, so sin(90) = 1. Hence, F = 2 C * 10 m/s * 0.5 T = 10 N.
Q5
Medium
Two point charges, +3 ฮผC and -3 ฮผC, are separated by a distance of 0.5 m in a vacuum. What is the electric field strength at the midpoint between the charges?
A.
0 N/C
B.
1.08 x 10^5 N/C
C.
3.6 x 10^5 N/C
D.
2.16 x 10^5 N/C
Show Answer & Explanation
Correct Answer: A
The electric fields produced by both charges at the midpoint will be equal in magnitude but opposite in direction, resulting in a net electric field of 0 N/C.
Q6
Medium
What is the expression for the gravitational potential energy (U) of an object of mass m at a distance r from a point mass M?
A.
-GMm/r
B.
GMm/r
C.
-GMm/r^2
D.
GMm/r^2
Show Answer & Explanation
Correct Answer: A
The correct formula for gravitational potential energy is U = -GMm/r. The negative sign indicates that gravitational force is attractive and potential energy decreases as the distance decreases.
Q7
Medium
A charged particle moves in a uniform electric field. What happens to the particle's kinetic energy as it moves from a region of higher electric potential to lower electric potential?
A.
It increases.
B.
It decreases.
C.
It remains constant.
D.
It becomes zero.
Show Answer & Explanation
Correct Answer: A
As the charged particle moves from a higher potential to a lower potential, it loses potential energy, which is transformed into kinetic energy, causing its kinetic energy to increase.
Q8
Hard
A charged particle is moving in a uniform magnetic field at an angle of 30 degrees to the field lines. If the particle has a charge of 2 ฮผC and a velocity of 5 m/s, what is the magnitude of the magnetic force acting on it? (Magnetic field strength is 0.1 T)
A.
0.1 N
B.
0.15 N
C.
0.2 N
D.
0.25 N
Show Answer & Explanation
Correct Answer: B
The magnetic force can be calculated using the formula F = qvB sin(ฮธ). Here, q = 2 ฮผC = 2 ร 10^-6 C, v = 5 m/s, B = 0.1 T, and ฮธ = 30ยฐ. Substituting these values gives F = (2 ร 10^-6) * (5) * (0.1) * sin(30ยฐ) = (2 ร 10^-6) * (5) * (0.1) * (0.5) = 0.0005 N = 0.15 N.
Q9
Hard
Two point charges, +3 ฮผC and -3 ฮผC, are separated by a distance of 0.5 m in a vacuum. What is the electric field at the midpoint between the charges?
A.
0 N/C
B.
6 ร 10^4 N/C towards the positive charge
C.
3 ร 10^4 N/C towards the negative charge
D.
6 ร 10^4 N/C towards the negative charge
Show Answer & Explanation
Correct Answer: A
At the midpoint, the electric fields due to the two charges are equal in magnitude but opposite in direction. Since the charges are equal and opposite, the total electric field at that point sums to zero, resulting in 0 N/C.
Q10
Hard
A charged particle moves in a uniform magnetic field. If the particle's velocity is perpendicular to the magnetic field, what will be the effect on the path of the particle?
A.
The particle will continue in a straight line.
B.
The particle will spiral outwards.
C.
The particle will move in a circular path.
D.
The particle will stop moving.
Show Answer & Explanation
Correct Answer: C
When a charged particle moves perpendicular to a magnetic field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path.
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Fields โ IB (International Baccalaureate) IB Physics HL Practice Questions Online
This page contains 148 practice MCQs for the chapter Fields in IB (International Baccalaureate) IB Physics HL. The questions are organized by difficulty โ 51 easy, 72 medium, 25 hard โ so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit.