Energy conservation - class-XII

Force on rod moving in a magnetic fileld is given by:

As the conductor falls, it cuts the horizontal component of the Earth's magnetic field. According to Fleming's Right Hand Rule, an induced EMF is generated. For a conductor falling and moving through the horizontal magnetic field, the current flows from West to East.

Magnetic field due to circular loop at center O due to current $i$ is given as,

Induced EMF is proportional to the rate of change of magnetic flux (dPhi/dt). Moving the magnet speedily increases the rate of change, resulting in a higher induced EMF. However, the total charge induced (q = delta Phi / R) depends only on the total change in flux, which is the same regardless of speed.

Given that,

Velocity, $v=\left( 3\hat{i}+4\hat{j}+\hat{k} \right)$

Electric field, $E=3\hat{i}+\hat{j}+2\hat{k}$

Magnetic field, $B=\hat{i}+\hat{j}+3\hat{k}$

Force,

$ F=q\left( \text{E+v}\times \text{B} \right) $

$ F=q\left( 3\hat{i}+\hat{j}+2\hat{k}+\left( (3\hat{i}+4\hat{j}+\hat{k})\times (\hat{i}+\hat{j}+3\hat{k} \right) \right) $

$ F=q\left( 3\hat{i}+\hat{j}+2\hat{k}+11\hat{i}-8\hat{j}-\hat{k} \right) $

$ F=q\left( 14\hat{i}-7\hat{j}+\hat{k} \right) $

So, y-component of force experienced by +q is

${{F} _{y}}=-7q$

$\begin{array}{l} \vec { B } , at, R, dis\tan  ce=\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2{ { \left( { { x^{ 2 } }+{ R^{ 2 } } } \right)  }^{ 3/2 } } } }  \ =\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2{ { \left( { 2{ R^{ 2 } } } \right)  }^{ 3/2 } } } }  \ =\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2.{ { \left( 8 \right)  }^{ 1/2 } }.{ R^{ 3 } } } }  \ =\frac { { { \mu _{ o } }i } }{ { 4\sqrt { 2 } R } }  \ \vec { B } , , at, , centre=\frac { { { \mu _{ o } }i } }{ { 2R } }  \ Ratio=\frac { { { \mu _{ o } }i.2R } }{ { 4\sqrt { 2 } R.{ \mu _{ o } }i } } =\frac { 1 }{ { 2\sqrt { 2 }  } } ={ \left( { \frac { 1 }{ 2 }  } \right) ^{ 3/2 } } \ Hence, \ option, , C, , is, , correct, , answer. \end{array}$

Magnetic induction on the equatorial line is proportional to 1/r^3. B1 / B2 = (r2 / r1)^3. (54 * 10^-6) / B2 = (0.3 / 0.2)^3 = (1.5)^3 = 3.375. B2 = (54 * 10^-6) / 3.375 = 16 * 10^-6 = 1.6 * 10^-5 T.

The induced emf in a rod moving in a magnetic field is given by the integral of (v x B) dot dl. For a wire carrying current I, the magnetic field at distance r is B = (mu_0 * I) / (2 * pi * r). Integrating this from r1 to r2 yields the formula in option B.

Magnetic induction B is given by B = mu * H, where mu is the permeability and H is the intensity of the magnetising field. Given mu = 3 * 10^-3 and H = 120 A/m, B = (3 * 10^-3) * 120 = 0.36 Wb/m^2.

Faraday's law states emf = -d(phi)/dt. The magnetic flux phi = B * A = (B_0 + alpha * t) * (pi * r^2). Differentiating with respect to time gives emf = -pi * r^2 * alpha.

The induced emf is given by E = (v x B) dot L. Here, v = 1 * i (m/s), B = 3i + 4j + 5k (T), and L = 5 * j (m). The cross product v x B = (1 * i) x (3i + 4j + 5k) = 4k - 5j. Then E = (4k - 5j) dot (5j) = -25V. The magnitude is 25V.

The change in flux is Delta(phi) = B * Delta(Area). The side increases from 0.5m to 0.5 + (0.1 * 4) = 0.9m. The area changes from 0.5^2 = 0.25 to 0.9^2 = 0.81. Delta(phi) = 0.2 * (0.81 - 0.25) = 0.2 * 0.56 = 0.112. Emf = Delta(phi) / Delta(t) = 0.112 / 4 = 0.028V.

In an electromagnetic wave, the relationship between electric field amplitude E_0 and magnetic field amplitude B_0 is E_0 = c * B_0, where c is the speed of light (3 * 10^8 m/s). E_0 = (3 * 10^8) * (510 * 10^-9) = 153 N/C.

The emf induced in a rotating rod is E = (1/2) * B * omega * L^2. Substituting B = 0.5 T, omega = 400 rad/s, and L = 1 m, we get E = 0.5 * 0.5 * 400 * 1^2 = 100 V.

Pitch x = v * cos(theta) * T = v * cos(theta) * (2 * pi * m) / (q * B). Radius R = (m * v * sin(theta)) / (q * B). Dividing R by x gives R / x = (sin(theta)) / (2 * pi * cos(theta)) = tan(60) / (2 * pi) = sqrt(3) / (2 * pi). Thus R = (sqrt(3) * x) / (2 * pi).

When a charged particle enters a magnetic field perpendicularly, the magnetic force acts perpendicular to the velocity. This force changes the direction of velocity but not its magnitude, meaning kinetic energy remains constant.

Mechanical power required $=$ electrical power expended due to the induced emf