Energy Stored in a Magnetic Field MCQ Quiz - Objective Question with Answer for Energy Stored in a Magnetic Field - Download Free PDF

Ans. (2) Sol.

Average energy density of magnetic field: U_m = m₀² / (2μ₀)

where m₀ is the maximum value of the magnetic field.

Average energy density of electric field: U_E = ε₀ε² / 2

Now, ε = C × m₀, where C² = 1 / (μ₀ × ε₀)

So, U_E = (ε₀ / 2) × C² × m₀²

Substituting the value of C²:

U_E = (ε₀ / 2) × (1 / (μ₀ × ε₀)) × m₀² = m₀² / (2μ₀) = U_m

Therefore, U_E = U_m

Since the energy densities of electric and magnetic fields are equal, the energy associated with equal volumes of each field is also equal. Hence, U_E / U_m = 1

Concept:

In an electromagnetic wave, the energy is stored in both electric and magnetic fields. The energy density of the electric field uE" id="MathJax-Element-346-Frame" role="presentation" style="position: relative;" tabindex="0">uEuE" id="MathJax-Element-187-Frame" role="presentation" style="position: relative;" tabindex="0">uE$u_E$ uE $u_E$ and the energy density of the magnetic field uB" id="MathJax-Element-347-Frame" role="presentation" style="position: relative;" tabindex="0">uBuB" id="MathJax-Element-188-Frame" role="presentation" style="position: relative;" tabindex="0">uB$u_B$ uB $u_B$ are given by:

$u_E=\frac{1}{2}{ϵ}_0{E}^2$

$u_B=\frac{1}{2}\frac{B^2}{{\mu }_0}$

where E" id="MathJax-Element-348-Frame" role="presentation" style="position: relative;" tabindex="0">EE" id="MathJax-Element-189-Frame" role="presentation" style="position: relative;" tabindex="0">E$E$ E $E$ is the electric field strength, B" id="MathJax-Element-349-Frame" role="presentation" style="position: relative;" tabindex="0">BB" id="MathJax-Element-190-Frame" role="presentation" style="position: relative;" tabindex="0">B$B$ B $B$ is the magnetic field strength, ϵ0" id="MathJax-Element-350-Frame" role="presentation" style="position: relative;" tabindex="0">ϵ0ϵ0" id="MathJax-Element-191-Frame" role="presentation" style="position: relative;" tabindex="0">ϵ0${ϵ}_0$ ϵ0 $\epsilon_0$ is the permittivity of free space, and μ0" id="MathJax-Element-351-Frame" role="presentation" style="position: relative;" tabindex="0">μ0μ0" id="MathJax-Element-192-Frame" role="presentation" style="position: relative;" tabindex="0">μ0${\mu }_0$ μ0 $\mu_0$ is the permeability of free space.

Explanation:

In an electromagnetic wave traveling in vacuum, the electric and magnetic fields are related by the speed of light c" id="MathJax-Element-352-Frame" role="presentation" style="position: relative;" tabindex="0">cc" id="MathJax-Element-193-Frame" role="presentation" style="position: relative;" tabindex="0">c$c$ c $c$ , where:

$E=cB$

Substituting E=cB" id="MathJax-Element-353-Frame" role="presentation" style="position: relative;" tabindex="0">E=cBE=cB" id="MathJax-Element-194-Frame" role="presentation" style="position: relative;" tabindex="0">E=cB$E=cB$ E=cB $E = cB$ into the energy densities, we get:

$u_E=\frac{1}{2}{ϵ}_0(cB{)}^2$

$u_B=\frac{1}{2}\frac{B^2}{{\mu }_0}$

Using the relation ϵ0μ0=1c2" id="MathJax-Element-354-Frame" role="presentation" style="position: relative;" tabindex="0">ϵ0μ0=1c2ϵ0μ0=1c2" id="MathJax-Element-195-Frame" role="presentation" style="position: relative;" tabindex="0">ϵ0μ0=1c2${ϵ}_0{\mu }_0=\frac{1}{c^2}$ ϵ0μ0=1c2 $\epsilon_0 \mu_0 = \frac{1}{c^2}$ , we can simplify the ratio of the energy densities:

$\frac{u_E}{u_B}=\frac{\frac{1}{2}{ϵ}_0(cB{)}^2}{\frac{1}{2}\frac{B^2}{{\mu }_0}}=\frac{{ϵ}_0{c}^2{B}^2}{\frac{B^2}{{\mu }_0}}=\frac{{ϵ}_0{c}^2}{\frac{1}{{\mu }_0}}=\frac{{ϵ}_0{c}^2{\mu }_0}{1}=\frac{1}{1}=1$

Therefore, the ratio of energy densities of electric and magnetic fields is 1:1.

The correct option is (1).

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

​F_B = 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

∵ F_B = qvBSinθ 

⇒ FB = qvBSin90º

⇒ FB = 2× 1.6× 10-19 × 105 × 0.80× 1

⇒ F_B = 2.56 × 10-14 N

∵ Option 3) is the right option.

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

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

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

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

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

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

Ans. (2) Sol.

Average energy density of magnetic field: U_m = m₀² / (2μ₀)

where m₀ is the maximum value of the magnetic field.

Average energy density of electric field: U_E = ε₀ε² / 2

Now, ε = C × m₀, where C² = 1 / (μ₀ × ε₀)

So, U_E = (ε₀ / 2) × C² × m₀²

Substituting the value of C²:

U_E = (ε₀ / 2) × (1 / (μ₀ × ε₀)) × m₀² = m₀² / (2μ₀) = U_m

Therefore, U_E = U_m

Since the energy densities of electric and magnetic fields are equal, the energy associated with equal volumes of each field is also equal. Hence, U_E / U_m = 1

Concept:

In an electromagnetic wave, the energy is stored in both electric and magnetic fields. The energy density of the electric field uE" id="MathJax-Element-346-Frame" role="presentation" style="position: relative;" tabindex="0">uEuE" id="MathJax-Element-187-Frame" role="presentation" style="position: relative;" tabindex="0">uE$u_E$ uE $u_E$ and the energy density of the magnetic field uB" id="MathJax-Element-347-Frame" role="presentation" style="position: relative;" tabindex="0">uBuB" id="MathJax-Element-188-Frame" role="presentation" style="position: relative;" tabindex="0">uB$u_B$ uB $u_B$ are given by:

$u_E=\frac{1}{2}{ϵ}_0{E}^2$

$u_B=\frac{1}{2}\frac{B^2}{{\mu }_0}$

where E" id="MathJax-Element-348-Frame" role="presentation" style="position: relative;" tabindex="0">EE" id="MathJax-Element-189-Frame" role="presentation" style="position: relative;" tabindex="0">E$E$ E $E$ is the electric field strength, B" id="MathJax-Element-349-Frame" role="presentation" style="position: relative;" tabindex="0">BB" id="MathJax-Element-190-Frame" role="presentation" style="position: relative;" tabindex="0">B$B$ B $B$ is the magnetic field strength, ϵ0" id="MathJax-Element-350-Frame" role="presentation" style="position: relative;" tabindex="0">ϵ0ϵ0" id="MathJax-Element-191-Frame" role="presentation" style="position: relative;" tabindex="0">ϵ0${ϵ}_0$ ϵ0 $\epsilon_0$ is the permittivity of free space, and μ0" id="MathJax-Element-351-Frame" role="presentation" style="position: relative;" tabindex="0">μ0μ0" id="MathJax-Element-192-Frame" role="presentation" style="position: relative;" tabindex="0">μ0${\mu }_0$ μ0 $\mu_0$ is the permeability of free space.

Explanation:

In an electromagnetic wave traveling in vacuum, the electric and magnetic fields are related by the speed of light c" id="MathJax-Element-352-Frame" role="presentation" style="position: relative;" tabindex="0">cc" id="MathJax-Element-193-Frame" role="presentation" style="position: relative;" tabindex="0">c$c$ c $c$ , where:

$E=cB$

Substituting E=cB" id="MathJax-Element-353-Frame" role="presentation" style="position: relative;" tabindex="0">E=cBE=cB" id="MathJax-Element-194-Frame" role="presentation" style="position: relative;" tabindex="0">E=cB$E=cB$ E=cB $E = cB$ into the energy densities, we get:

$u_E=\frac{1}{2}{ϵ}_0(cB{)}^2$

$u_B=\frac{1}{2}\frac{B^2}{{\mu }_0}$

Using the relation ϵ0μ0=1c2" id="MathJax-Element-354-Frame" role="presentation" style="position: relative;" tabindex="0">ϵ0μ0=1c2ϵ0μ0=1c2" id="MathJax-Element-195-Frame" role="presentation" style="position: relative;" tabindex="0">ϵ0μ0=1c2${ϵ}_0{\mu }_0=\frac{1}{c^2}$ ϵ0μ0=1c2 $\epsilon_0 \mu_0 = \frac{1}{c^2}$ , we can simplify the ratio of the energy densities:

$\frac{u_E}{u_B}=\frac{\frac{1}{2}{ϵ}_0(cB{)}^2}{\frac{1}{2}\frac{B^2}{{\mu }_0}}=\frac{{ϵ}_0{c}^2{B}^2}{\frac{B^2}{{\mu }_0}}=\frac{{ϵ}_0{c}^2}{\frac{1}{{\mu }_0}}=\frac{{ϵ}_0{c}^2{\mu }_0}{1}=\frac{1}{1}=1$

Therefore, the ratio of energy densities of electric and magnetic fields is 1:1.

The correct option is (1).

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

​F_B = 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

∵ F_B = qvBSinθ 

⇒ FB = qvBSin90º

⇒ FB = 2× 1.6× 10-19 × 105 × 0.80× 1

⇒ F_B = 2.56 × 10-14 N

∵ Option 3) is the right option.