Magnetic Force - Definition of B MCQ Quiz in मल्याळम - Objective Question with Answer for Magnetic Force - Definition of B - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

 Concept:

• Mutual Inductance is a measure of the ability of one circuit to induce an electromotive force (EMF) in another circuit through a change in current.
• The SI unit of mutual inductance is the Henry (H).
• Magnetic Flux (Φ) is the total magnetic field passing through a given area.
• The SI unit of magnetic flux is the Weber (Wb).
• The mutual inductance M between two loops is given by: M = (N₂Φ₂) / I₁,

where Φ₂ = magnetic flux through the second loop due to the current I₁ in the first loop.

• When a current I flows through a loop of wire, it produces a magnetic field B at a distance from the loop, and the magnetic field strength at the center of a square loop of side length L is given by:
• B = (μ₀I) / (2L),

Where μ₀ = permeability of free space.

Calculation:

Here,
Side of small loop = l
Side of large loop = L
Current in large loop = i
Permeability of free space = μ₀

We know, ϕ = Mi

⇒ \(\phi \simeq\left[4 \times \frac{\mu_0 i}{4 \pi(L / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]\right] \times \text { Area }\)

= \(\frac{2 \sqrt{2} \mu_0 i}{\pi L} \times I^2\)

⇒ \(M=\frac{\phi}{i}=\frac{2 \sqrt{2} \mu_0 I^2}{\pi L}\)

The mutual inductance of the system is M = \(\frac{2 \sqrt{2} \mu_0 I^2}{\pi L}\)

∴ The correct option is 3

CONCEPT:

→The magnetic field exerted on the electron is written as;

\(B=\frac{E}{c}\)

Here we have B as the magnetic field, E as the electric field, and c as the velocity of light.

→The maximum force exerted is written as follows;

\(F _{max}=eBv\)

Here we have B as the magnetic field, v is the speed and e is the electron.

CALCULATION:

Given: E = 800 sin ω \(\left( {{\rm{t}}\,{\rm{ - }}\,\frac{{\rm{x}}}{{\rm{c}}}} \right)\)

\(v = 3 \times 10^7\)

As we know,

\(B=\frac{E}{c}\)

Now, on putting the values we have;

\(B=\frac{800 sin \omega (t - \frac{x}{c})}{c}\)

⇒ \(B=\frac{800 }{3 \times 10 ^8}\)

As we know,

\(F _{max}=eBv\)

⇒ \(F _{max}=1.6 \times 10^{-19} \times \frac{800}{3 \times 10^8} \times 3 \times 10^7\)

⇒ \(F _{max}=12.8 \times 10^{-18} \) N

Hence, option 2) is the correct answer.

Concept:

The magnetic field produced by a steady current flowing in a very long straight wire. The magnitude of the magnetic field is

\(B = \frac{{{\mu _0}I}}{{2\pi R}}\)

• Parallel wires carrying currents will exert forces on each other.
• Each wire produces a magnetic field, which influences the other wire. 
• When the currents in both wires flow in the same direction, then the force is attractive. 
• When the currents flow in opposite directions, then the force is repulsive.

The force exerted on wires can be given by:

\(F= \frac{{{\mu _0}}}{{2\pi }}\frac{{{I_1}{I_2}}}{{{r_1}}} \times Length\;of\;wire\)

Calculation:

Repulsion by wire D,       

\({F_1} = \frac{{{\mu _0}}}{{2\pi }}\frac{{{i_1}{i_2}l}}{r}\)   [towards right]

= \(\frac{{\left( {2 \;\times\; {{10}^{ - 7}}} \right)\left( {30\; \times\; 10} \right)}}{{3\; \times\; {{10}^{ - 2}}}}\left( {0.25} \right)\)

= 5 × 10^-4 N

Repulsion by wire G,

\({F_2} = \frac{{\left( {2\; \times\; {{10}^{ - 7}}} \right)\left( {20\; \times\; 10} \right)}}{{5\; \times \;{{10}^{ - 2}}}}\left( {0.25} \right)\)   [towards left]

= 2 × 10^-4 N

∴ F_net = F₁ – F₂

= 3 × 10^-4 N  [towards right]

Calculation:

Initially, forces acting on the rod-ring system are weight mg and tension 2T0. In equilibrium, mg = 2T0.

The ring takes time T = 2π/ω to complete one revolution, setting a current i and magnetic moment μ, given by:

i = Q / T = Qω / (2π),    μ = iA = (1/2) ω Q R2

The direction of μ is towards the right. The torque on the loop is τ = |μ × B| = (1/2) ω B Q R2 (anticlockwise).

To balance this torque, tensions in the strings change to new values T1 and T2 such that T1 > T2. In equilibrium, the resultant force and resultant torque about the centre of the disc are zero.

T1 + T2 = mg        (1)

T₁(D/2) - T₂(D/2) = (1/2) ωBQR²       (2)

from (1) and (2) we get

T₁ = T₀ + ωBQR²/ (2D) 

⇒ ω_max = (DT₀)/ (BQR²)

Thus by comparing we have α - β is -1 +2 = 1.

Calculation:
For the semicircle KNM lying in the x-z plane, take a small element at angle θ. The force on a current element is dF = I × (dl × B).

Calculating and integrating over the arc, we find the force F₂ = 2IB₀R î.

For the semicircle KLM in the y-z plane, again consider a small element at angle θ.

Following the same procedure, the total force on this arc is F₁ = 2IB₀R î.

The total force on the loop is:

F = F₁ + F₂ = 4IB₀R î

 F₁ = F₂ = 2IB₀R î, so net force F = 4IB₀R î

Calculation:

A → B

\(\overrightarrow{\mathrm{V}}=-v_{x} \hat{\mathrm{i}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}\)

\(\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}(-\hat{\mathrm{k}})\)

\(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{r}}\left[-\mathrm{v}_{\mathrm{x}} \hat{\mathrm{j}}-\mathrm{v}_{\mathrm{y}} \hat{\mathrm{i}}\right]\)

\(\mathrm{a}_{\mathrm{x}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \cdot \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}\)

\(\mathrm{a}_{\mathrm{y}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \cdot \frac{\mathrm{v}_{\mathrm{x}}}{\mathrm{r}}\)

\(\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{dr}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}\)

\(\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{v}_{\mathrm{y}}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{dr}}{\mathrm{r}}\)

\(\int_{0}^{\mathrm{v}_{0}} \frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\sqrt{\mathrm{v}_{0}^{2}-\mathrm{v}_{\mathrm{x}}^{2}}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \int_{\mathrm{a}}^{\mathrm{x}_{1}} \frac{\mathrm{dr}}{\mathrm{r}}\)

Let, z² = v₀² – v_x²

2zdz = –2v_x dv_x

zdz = – v_x dv_x

\(\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\sqrt{\mathrm{v}_{0}^{2}-\mathrm{v}_{\mathrm{x}}^{2}}}=\frac{-\mathrm{zdz}}{\mathrm{z}}=-\mathrm{dz}\)

then integral becomes 

\(-\int_{\mathrm{v}_{0}}^{0} \mathrm{dz}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}_{1}}{\mathrm{a}}\)

\(\mathrm{v}_{0}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}_{1}}{\mathrm{a}}\)

\(\mathrm{X}_{1}=\mathrm{a} \mathrm{e}^{-\frac{2 \pi \mathrm{mv}_{0}}{\mu_{0} \mathrm{Iq}}} \ldots \ldots .(1)\)

For B → C 

\(\overrightarrow{\mathrm{v}}=-v_{x} \hat{\mathrm{i}}-v_{y} \hat{\mathrm{j}}\)

\(\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}(-\hat{\mathrm{k}})\)

\(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{r}}\left(-\mathrm{v}_{\mathrm{x}} \hat{\mathrm{j}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{i}}\right)\)

\(\mathrm{a}_{\mathrm{x}}=+\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}} \quad \mathrm{a}_{\mathrm{y}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \cdot \frac{\mathrm{v}_{\mathrm{x}}}{\mathrm{r}}\)

\(\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{dr}}=\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}\)

\(\int_{v_{0}}^{0} \frac{v_{x} \mathrm{dv}_{x}}{\sqrt{v_{0}^{2}-v_{x}^{2}}}=\frac{\mu_{0} I \mathrm{Iq}}{2 \pi m} \int_{x_{1}}^{\mathrm{x}} \frac{\mathrm{dr}}{\mathrm{r}}\)

\(\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}}{\mathrm{x}_{1}}=-\int_{0}^{\mathrm{v}_{0}} \mathrm{~d} \mathrm{z}=-\mathrm{v}_{0}\)

\(X=X_{1} e^{-\frac{2 \pi m v_{0}}{\mu_{0} \mathrm{Iq}}} \ldots \ldots(2)\)

From equation 1 and 2 

\(X=a e^{-\frac{4 \pi m v_{0}}{\mu_{0} \mathrm{Iq}}}\)

 Concept:

• Mutual Inductance is a measure of the ability of one circuit to induce an electromotive force (EMF) in another circuit through a change in current.
• The SI unit of mutual inductance is the Henry (H).
• Magnetic Flux (Φ) is the total magnetic field passing through a given area.
• The SI unit of magnetic flux is the Weber (Wb).
• The mutual inductance M between two loops is given by: M = (N₂Φ₂) / I₁,

where Φ₂ = magnetic flux through the second loop due to the current I₁ in the first loop.

• When a current I flows through a loop of wire, it produces a magnetic field B at a distance from the loop, and the magnetic field strength at the center of a square loop of side length L is given by:
• B = (μ₀I) / (2L),

Where μ₀ = permeability of free space.

Calculation:

Here,
Side of small loop = l
Side of large loop = L
Current in large loop = i
Permeability of free space = μ₀

We know, ϕ = Mi

⇒ \(\phi \simeq\left[4 \times \frac{\mu_0 i}{4 \pi(L / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]\right] \times \text { Area }\)

= \(\frac{2 \sqrt{2} \mu_0 i}{\pi L} \times I^2\)

⇒ \(M=\frac{\phi}{i}=\frac{2 \sqrt{2} \mu_0 I^2}{\pi L}\)

The mutual inductance of the system is M = \(\frac{2 \sqrt{2} \mu_0 I^2}{\pi L}\)

∴ The correct option is 3

Calculation:
A magnetic field does not exert a force on a stationary electric charge. The magnetic component of the Lorentz force is proportional to the velocity of the charge. If the charge is at rest, the velocity is zero, so the magnetic force is zero. The Lorentz force is given by:

F = q(E + v × B)

Since v = 0 for a stationary charge, the magnetic force component (v × B) is zero, and a magnetic field does not interact with a stationary electric charge.

Concept:

The force on a current-carrying conductor in a magnetic field is given by:

\(F = I \cdot (L \times B)\)

Where:

• I: Current through the conductor (in amperes).
• L: Length of the conductor (in meters).
• B: Magnetic field strength (in tesla).

For a semicircular conductor of radius r, the length of the conductor is given by \(L = \pi r.\)

Force is distributed equally between the two springs, and hence, each spring experiences half of the total force.

Calculation:

Given:

• Current through the wire, I.
• Radius of the semicircular wire, r.
• Magnetic field strength, B.

Using the formula for the total force:

\(⇒ F = I \cdot L \cdot B\\ ⇒ F = I \cdot (\pi r) \cdot B\)

Force on each spring is:

\(⇒ T = \frac{F}{2}\\ ⇒ T = \frac{I \cdot (\pi r) \cdot B}{2} \\ ⇒ T = I \cdot r \cdot B\)

∴ The force acting on each spring is\( I \cdot r \cdot B\).

The correct option is 3).

Concept:  

To solve the problem, we need to balance the forces acting on the rod.

Gravitational force component along the incline:

Fg = mgsinθ 

Magnetic force

F_m = ILBsinθ 

Calculation: 

Given:

Linear density of the rod λ = 0.45kg/m

Magnetic field: B = 0.15 T

Angle of the incline θ = 45

Acceleration due to gravity: g = 10m/s²

The mass per unit length of the rod is represented by m = λ.L 

For the rod to remain stationary, the magnetic force must balance the component of the gravitational force along the incline:

⇒ ILBsinθ  = mgsinθ 

⇒ ILB = λLg

⇒ I = \(\frac{\lambda.g}{B}\)

⇒ I = \(\frac{0.45 \times 10}{0.15} = 30\)

∴ the minimum current required to keep the rod stationary is 30 A.