Energy Stored in a Magnetic Field MCQ Quiz - Objective Question with Answer for Energy Stored in a Magnetic Field - Download Free PDF
Last updated on May 23, 2025
Latest Energy Stored in a Magnetic Field MCQ Objective Questions
Energy Stored in a Magnetic Field Question 1:
If Uɛ electric field energy density and Um magnetic field energy density. Then
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 1 Detailed Solution
Ans. (2) Sol.
Average energy density of magnetic field: Um = m₀² / (2μ₀)
where m₀ is the maximum value of the magnetic field.
Average energy density of electric field: UE = ε₀ε² / 2
Now, ε = C × m₀, where C² = 1 / (μ₀ × ε₀)
So, UE = (ε₀ / 2) × C² × m₀²
Substituting the value of C²:
UE = (ε₀ / 2) × (1 / (μ₀ × ε₀)) × m₀² = m₀² / (2μ₀) = Um
Therefore, UE = Um
Since the energy densities of electric and magnetic fields are equal, the energy associated with equal volumes of each field is also equal. Hence, UE / Um = 1
Energy Stored in a Magnetic Field Question 2:
In an electromagnetic wave, the ratio of energy densities of electric and magnetic fields is ____________. Fill in the blank with the correct answer from the options given below.
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 2 Detailed Solution
Concept:
In an electromagnetic wave, the energy is stored in both electric and magnetic fields. The energy density of the electric field
\(u_E = \frac{1}{2} \epsilon_0 E^2\)
\(u_B = \frac{1}{2} \frac{B^2}{\mu_0}\)
where
Explanation:
In an electromagnetic wave traveling in vacuum, the electric and magnetic fields are related by the speed of light
\(E = cB\)
Substituting
\(u_E = \frac{1}{2} \epsilon_0 (cB)^2\)
\(u_B = \frac{1}{2} \frac{B^2}{\mu_0}\)
Using the relation
\( \frac{u_E}{u_B} = \frac{\frac{1}{2} \epsilon_0 (cB)^2}{\frac{1}{2} \frac{B^2}{\mu_0}} = \frac{\epsilon_0 c^2 B^2}{\frac{B^2}{\mu_0}} = \frac{\epsilon_0 c^2}{\frac{1}{\mu_0}} = \frac{\epsilon_0 c^2 \mu_0}{1} = \frac{1}{1} = 1\)
Therefore, the ratio of energy densities of electric and magnetic fields is 1:1.
The correct option is (1).
Energy Stored in a Magnetic Field Question 3:
A He2+ ion travels at right angles to a magnetic field of 0.80 T with a velocity of 105 m/s. The magnitude of the magnetic force on the ion will be : (electron charge = 1.6 × 10-19 C)
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 3 Detailed Solution
Concept:
- Magnetic field intensity - is a measure of how strong or weak any magnetic field is.
- The SI unit of B is Ns/(Cm) = Tesla (T)
- Lorentz Force - The force a magnetic field exerted on a charge q moving with velocity v is called magnetic Lorentz force and it is represented by
FB = qvBSinθ.
Given:
Magnetic field intensity(B)= 0.80T
Particle = He2+
Angle of interaction(θ) = 90º
Charge on particle(q) = 1.6× 10-19
Velocity of particle(v)= 105 m/s
∵ FB = qvBSinθ
⇒ FB = qvBSin90º
⇒ FB = 2× 1.6× 10-19 × 105 × 0.80× 1
⇒ FB = 2.56 × 10-14 N
∵ Option 3) is the right option.
Energy Stored in a Magnetic Field Question 4:
If 10A current is flowing through a solenoid of inductance 5H, find the magnetic energy stored in the solenoid.
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 4 Detailed Solution
CONCEPT:
- Solenoid: It is a coil wound into a tightly packed helix, that generates a controlled magnetic field.
- The uniform magnetic field in the solenoid is produced when an electric current is passed through it.
- Inductor: A device or component of a circuit that has significant self-inductance and stores energy in a magnetic field.
The formula used to find the energy stored in a magnetic field is given by:
E = 1/2 × L × I2
Where L is the inductance and I is current in the solenoid.
CALCULATION:
given that L = 5H; current I = 10A
The magnetic energy stored in the solenoid E = 1/2 × L × I2
E = 1/2 × 5 × 102 = 250 Joule
- So the correct answer will be option 2 i.e. 250 J
Energy Stored in a Magnetic Field Question 5:
A solenoid of inductance 2H carries a current of 1 A. what is the magnetic energy stored in a solenoid?
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 5 Detailed Solution
CONCEPT:
- Solenoid: A type of electromagnet that generates a controlled magnetic field through a coil wound into a tightly packed helix.
- The uniform magnetic field is produced when an electric current is passed through it.
- Inductor: A device or component of a circuit that has significant self-inductance and stores energy in a magnetic field.
- The formula used to find the energy stored in a magnetic field is
E = 1/2 × L × I2
Where L is the inductance and I is current in the solenoid.
CALCULATION:
given that L = 2H; current I = 1A
magnetic energy stored in the solenoid E = 1/2 × L × I2
E = 1/2 × 2 × 12 = 1 Joule
- So the correct answer will be option 2 i.e. 1 J
Top Energy Stored in a Magnetic Field MCQ Objective Questions
If 10A current is flowing through a solenoid of inductance 5H, find the magnetic energy stored in the solenoid.
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 6 Detailed Solution
Download Solution PDFCONCEPT:
- Solenoid: It is a coil wound into a tightly packed helix, that generates a controlled magnetic field.
- The uniform magnetic field in the solenoid is produced when an electric current is passed through it.
- Inductor: A device or component of a circuit that has significant self-inductance and stores energy in a magnetic field.
The formula used to find the energy stored in a magnetic field is given by:
E = 1/2 × L × I2
Where L is the inductance and I is current in the solenoid.
CALCULATION:
given that L = 5H; current I = 10A
The magnetic energy stored in the solenoid E = 1/2 × L × I2
E = 1/2 × 5 × 102 = 250 Joule
- So the correct answer will be option 2 i.e. 250 J
A solenoid of inductance 2H carries a current of 1 A. what is the magnetic energy stored in a solenoid?
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 7 Detailed Solution
Download Solution PDFCONCEPT:
- Solenoid: A type of electromagnet that generates a controlled magnetic field through a coil wound into a tightly packed helix.
- The uniform magnetic field is produced when an electric current is passed through it.
- Inductor: A device or component of a circuit that has significant self-inductance and stores energy in a magnetic field.
- The formula used to find the energy stored in a magnetic field is
E = 1/2 × L × I2
Where L is the inductance and I is current in the solenoid.
CALCULATION:
given that L = 2H; current I = 1A
magnetic energy stored in the solenoid E = 1/2 × L × I2
E = 1/2 × 2 × 12 = 1 Joule
- So the correct answer will be option 2 i.e. 1 J
Energy Stored in a Magnetic Field Question 8:
If 10A current is flowing through a solenoid of inductance 5H, find the magnetic energy stored in the solenoid.
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 8 Detailed Solution
CONCEPT:
- Solenoid: It is a coil wound into a tightly packed helix, that generates a controlled magnetic field.
- The uniform magnetic field in the solenoid is produced when an electric current is passed through it.
- Inductor: A device or component of a circuit that has significant self-inductance and stores energy in a magnetic field.
The formula used to find the energy stored in a magnetic field is given by:
E = 1/2 × L × I2
Where L is the inductance and I is current in the solenoid.
CALCULATION:
given that L = 5H; current I = 10A
The magnetic energy stored in the solenoid E = 1/2 × L × I2
E = 1/2 × 5 × 102 = 250 Joule
- So the correct answer will be option 2 i.e. 250 J
Energy Stored in a Magnetic Field Question 9:
A solenoid of inductance 2H carries a current of 1 A. what is the magnetic energy stored in a solenoid?
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 9 Detailed Solution
CONCEPT:
- Solenoid: A type of electromagnet that generates a controlled magnetic field through a coil wound into a tightly packed helix.
- The uniform magnetic field is produced when an electric current is passed through it.
- Inductor: A device or component of a circuit that has significant self-inductance and stores energy in a magnetic field.
- The formula used to find the energy stored in a magnetic field is
E = 1/2 × L × I2
Where L is the inductance and I is current in the solenoid.
CALCULATION:
given that L = 2H; current I = 1A
magnetic energy stored in the solenoid E = 1/2 × L × I2
E = 1/2 × 2 × 12 = 1 Joule
- So the correct answer will be option 2 i.e. 1 J
Energy Stored in a Magnetic Field Question 10:
If Uɛ electric field energy density and Um magnetic field energy density. Then
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 10 Detailed Solution
Ans. (2) Sol.
Average energy density of magnetic field: Um = m₀² / (2μ₀)
where m₀ is the maximum value of the magnetic field.
Average energy density of electric field: UE = ε₀ε² / 2
Now, ε = C × m₀, where C² = 1 / (μ₀ × ε₀)
So, UE = (ε₀ / 2) × C² × m₀²
Substituting the value of C²:
UE = (ε₀ / 2) × (1 / (μ₀ × ε₀)) × m₀² = m₀² / (2μ₀) = Um
Therefore, UE = Um
Since the energy densities of electric and magnetic fields are equal, the energy associated with equal volumes of each field is also equal. Hence, UE / Um = 1
Energy Stored in a Magnetic Field Question 11:
In an electromagnetic wave, the ratio of energy densities of electric and magnetic fields is ____________. Fill in the blank with the correct answer from the options given below.
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 11 Detailed Solution
Concept:
In an electromagnetic wave, the energy is stored in both electric and magnetic fields. The energy density of the electric field
\(u_E = \frac{1}{2} \epsilon_0 E^2\)
\(u_B = \frac{1}{2} \frac{B^2}{\mu_0}\)
where
Explanation:
In an electromagnetic wave traveling in vacuum, the electric and magnetic fields are related by the speed of light
\(E = cB\)
Substituting
\(u_E = \frac{1}{2} \epsilon_0 (cB)^2\)
\(u_B = \frac{1}{2} \frac{B^2}{\mu_0}\)
Using the relation
\( \frac{u_E}{u_B} = \frac{\frac{1}{2} \epsilon_0 (cB)^2}{\frac{1}{2} \frac{B^2}{\mu_0}} = \frac{\epsilon_0 c^2 B^2}{\frac{B^2}{\mu_0}} = \frac{\epsilon_0 c^2}{\frac{1}{\mu_0}} = \frac{\epsilon_0 c^2 \mu_0}{1} = \frac{1}{1} = 1\)
Therefore, the ratio of energy densities of electric and magnetic fields is 1:1.
The correct option is (1).
Energy Stored in a Magnetic Field Question 12:
A He2+ ion travels at right angles to a magnetic field of 0.80 T with a velocity of 105 m/s. The magnitude of the magnetic force on the ion will be : (electron charge = 1.6 × 10-19 C)
Answer (Detailed Solution Below)
Energy Stored in a Magnetic Field Question 12 Detailed Solution
Concept:
- Magnetic field intensity - is a measure of how strong or weak any magnetic field is.
- The SI unit of B is Ns/(Cm) = Tesla (T)
- Lorentz Force - The force a magnetic field exerted on a charge q moving with velocity v is called magnetic Lorentz force and it is represented by
FB = qvBSinθ.
Given:
Magnetic field intensity(B)= 0.80T
Particle = He2+
Angle of interaction(θ) = 90º
Charge on particle(q) = 1.6× 10-19
Velocity of particle(v)= 105 m/s
∵ FB = qvBSinθ
⇒ FB = qvBSin90º
⇒ FB = 2× 1.6× 10-19 × 105 × 0.80× 1
⇒ FB = 2.56 × 10-14 N
∵ Option 3) is the right option.