Magnetic Force Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. What is the force experienced by a magnet at the poles if the north pole of that magnet is brought near a stationary negatively charged conductor?

1. Zero
2. Maximum
3. Depends on the nature of the conductor
4. None of the above

Ans. The correct answer is a. Zero

Explanation: Because the stationary charge generates no magnetic field, the north pole of the magnet will experience no force. As a result, the magnet will experience zero force at the poles.

Ques. On a moving charged particle, what will be the work done by the magnetic field?

1. Depends on the strength of the magnetic field
2. Zero
3. Maximum
4. Minimum

Ans. The correct answer is b. Zero

Explanation: Because the magnetic force works in a direction perpendicular to the direction of motion or velocity, the work done will be zero.

W = F x dl x cos 90° = 0

Ques. The magnetic effect that can be experienced in the space around a current-carrying conductor is called

1. Magnetic field
2. Current
3. Electric pole
4. Magnetic flux

Ans. The correct answer is a. Magnetic field

Explanation: The magnetic field is the region around a current-carrying conductor where its magnetic influence may be felt. When a current is conducted through a conductor, it alters the space surrounding it, resulting in the formation of a magnetic field.

Ques. Which of the following conditions shows that if a charge moves through a uniform magnetic field, then the force acting on it is maximum?

1. θ = 100°
2. θ = 30°
3. θ = 60°
4. θ = 90°

Ans. The correct answer is d. θ = 100°

Explanation: The magnetic force (Fm) is maximum when θ = 90°, hence this is the state in which the force experienced is greatest. This means the charge suffers the most force while moving to the magnetic field's direction at a right angle.

Ques. The magnetic field outside a solenoid will be

1. Triple the value of the field inside
2. Double the value of the field inside
3. Zero
4. Infinity

Ans. The correct answer is c. Zero

Explanation: Because there are no magnetic lines of force outside a solenoid, the magnetic field will be zero.

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Short Answers Questions [2 Marks Questions]

Ques. Define Magnetic force.

Ans. Magnetic force is defined as the attractive or repulsive force exerted between a magnet's poles and electrically charged moving particles. As a result, it is a result of electromagnetic forces.

Ques. Define electromagnetic force.

Ans. The electromagnetic force is one of nature's four basic forces. It is the most dominant force in atom-molecule interactions. Electromagnetic forces are found between two charged particles, attracting particles with opposing charges and repelling particles with the same charge. Many chemical and physical occurrences observed in everyday life are caused by electromagnetic force. Atoms are held together by the electrical interaction between their nuclei and electrons.

Ques. Define magnetic field.

Ans. The region or space around the current-carrying conductor or a magnet within which its influence can be felt by the magnetic needle is called a magnetic field.

The strength of the magnetic field is also known as magnetic flux density, magnetic induction, or magnetic vector.

Ques. What is Fleming’s left-hand rule?

Ans. According to Fleming’s left-hand rule, stretch the left hand such that the fore-finger, the central, and the thumb are mutually perpendicular to each other. When the forefinger points in the direction of the magnetic field and the central finger points in the direction of current, then the thumb gives the direction of force acting on the conductor.

Also Read:

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Long Answers Questions [3 Marks Questions]

Ques. An 8 cm wire with a current of 2 A is oriented 36° from parallel to a magnetic field with a strength of 6 T. What is the force on the wire?

Ans. Given

• The length of the wire, L = 8 cm = 8 x 10^-2 m
• The current flowing through the wire, I = 2 A
• The strength of the magnetic field, B = 6 T
• The angle between the direction of the current element (wire) and magnetic field, θ = 36°

The magnetic force acting on a current-carrying wire in a magnetic field is given by

F = BIL sinθ

On substituting the values, we get

F = 6 x 2 x 8 x 10^-2 x sin 36°

⇒ F = 96 x 10^-2 x 0.5878

⇒ F = 56.4 x 10^-2 = 0.564 N

Hence the force acting on the wire is 0.564 N.

Ques. A long conductor of length 1.5 m is placed parallel to a short conductor of length 5 cm near its center. The conductors are carrying currents of 3 A and 4 A respectively in the same direction. Calculate the total force experienced by a short conductor, when the distance between them is 0.03 m.

Ans. Given

• Current flowing through the long conductor, I₁ = 3 A
• Current flowing through the short conductor, I₂ = 4 A
• Distance between the conductors, d = 0.03 m
• Length of the long conductor, L₁ = 1.5 m
• Length of the shortest conductor, L₂ = 5 cm = 5 x 10^-2 m

The magnetic force per unit length between the two parallel straight conductors is given by

\(\frac{F}{L} = \frac{\mu _0}{4 \pi} \frac{I_1I_2}{d}\)

On substituting the values, the magnetic force per unit length experienced by the short conductor is given by

 \(\frac{F}{L_2} = 10^{-7} \times \frac{3 \times 2}{0.03} = 8 \times 10^{-5} N\)

Therefore the magnetic force experienced by the short conductor is given by

F = 8 x 10^-5 x L²

⇒ F = 8 x 10^-5 x 5 x 10^-2 = 4 x 10^-6 N

Ques. A solenoid is 2 m long and 3 cm in diameter. It has 5 layers of windings of 1000 turns each and carries a current of 5 A. What is the magnetic field at its center?

Ans. The magnetic field inside a solenoid is given by

\(B = \frac{\mu _0NI}{L}\)

Where

• Absolute permeability of free space, µ₀ = 4π x 10^-7 H/m
• N is the number of turns
• I is the current flowing through the solenoid
• L is the length of the solenoid

Given

• N = 1000 turns
• I = 5 A
• L = 2 m

On substituting the values, we get

B = (4π x 10^-7 x 1000 x 5) / 2 = 0.314 x 10^-2 T

Since the solenoid has 5 layers of windings of same number of turns, therefore magnetic field inside the solenoid is

B’ = 5 x 0.314 x 10^-2 = 1.57 x 10^-2 T

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Very Long Answers Questions [5 Marks Questions]

Ques. A magnetic field of 5.0 x 10^-4 T just balances a perpendicular electric field of 15 kVm^-1 in their effect on an electron beam passing through the two fields in a direction perpendicular to both of them. What is the speed of the electrons?

Ans. Given

• The magnitude of the magnetic field, B = 5.0 x 10^-4 T
• The magnitude of the electric field, E = 15 kVm^-1 = 15 x 10³ Vm^-1

Since the magnetic field balances the electric field, therefore force due to the magnetic field (F_m) will be equal to the force due to the electric field (F_e) i.e.

F_m = F_e …(i)

Now force due to the magnetic field is given by

F_m = qvB

Ans force due to the electric field is given by

F_e = qE

Therefore from equation (i), we get

qvB = qE

⇒ v = E/B

Where v is the speed of the electron, Hence on substituting the values, we get

v = (15 x 10³) / (5 x 10^-4) = 3 x 10⁷ m/s

Ques. A particle with a charge of 2 µC is moving at 3 x 10⁶ m/s perpendicularly through a magnetic field with a strength of 0.05 T. What is the magnitude of the force on the particle?

Ans. Given

• Charge on the particle, q = 2 µC = 2 x 10^-6 C
• The velocity of the charged particle, v = 3 x 10⁶ m/s
• The magnitude of the magnetic field, B = 0.05 T

The magnetic force experienced by a charged particle moving in a magnetic field is given by

F = qvB sinθ

Where θ is the angle between the velocity vector of the charged particle and the direction of the magnetic field.

Since the charged particle is moving perpendicularly through the magnetic field, therefore, θ = 90°

On substituting the values, we get

F = 2 x 10^-6 x 3 x 10⁶ x 0.05 x sin 90°

⇒ F = 0.3 x 1 = 0.3 N

Hence the magnitude of the force acting on the particle is 0.3 N

Ques. A horizontal component of Earth’s magnetic field at a certain plane is 3.0 x 10^-5 T and the direction of the field is from the geographical south to the geographical north. A very long straight conductor is carrying a steady current of 1 A. What is the force per unit length on it when it is placed on a horizontal table and the direction of current is

• East to West
• South to North

Ans. Given

• The magnitude of the Earth’s magnetic field, B = 3.0 x 10^-5 T
• The current flowing through the conductor, I = 1 A

• When the direction of the current is from east to west, then the angle between the current element and the magnetic field is θ = 90°

Using the formula of magnetic force F = BIL sinθ, we have

Force per unit length, F/L = BI sinθ

On substituting the values, we get

F/L = 3.0 x 10^-5 x 1 x sin90°

⇒ F/L = 3 x 10^-5 N/m

• When the direction of the current is from south to north, then the angle between the current element and the magnetic field is θ = 0°

Using the formula of magnetic force F = BIL sinθ, we have

Force per unit length, F/L = BI sinθ

On substituting the values, we get

F/L = 3.0 x 10^-5 x 1 x sin0°

⇒ F/L = 0

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