CBSE Class 12 Physics Chapter 11: Electromagnetic Induction NCERT Solutions

Solution:

The problem describes a stationary rectangular loop experiencing a decreasing magnetic field. We need to calculate the power dissipated as heat and identify the source of this power.

First, let's determine the area of the rectangular loop. The sides are given as 8 cm and 2 cm.

Area, $A\; =\; \backslash text\{length\}\; \backslash times\; \backslash text\{width\}$

$A\; =\; 8\; \backslash text\{\; cm\}\; \backslash times\; 2\; \backslash text\{\; cm\}\; =\; 16\; \backslash text\{\; cm\}^2$

To use this in calculations involving SI units, we convert the area to square meters:

$A\; =\; 16\; \backslash times\; 10^\{-4\}\; \backslash text\{\; m\}^2$

The initial magnetic field is given as $B\; =\; 0.3\; \backslash text\{\; T\}$, but this value is not directly needed for calculating the induced EMF if the rate of change is given.

The rate at which the magnetic field is decreasing is given as:

$\backslash frac\{dB\}\{dt\}\; =\; -0.02\; \backslash text\{\; T/s\}$ (The negative sign indicates a decrease).

According to Faraday's law of electromagnetic induction, the magnitude of the induced EMF ($e$) in the loop is equal to the rate of change of magnetic flux ($\backslash phi$) through it:

$e\; =\; \backslash left|\; \backslash frac\{d\backslash phi\}\{dt\}\; \backslash right|$

The magnetic flux is given by $\backslash phi\; =\; AB\; \backslash cos\backslash theta$. Since the loop is stationary and the magnetic field is perpendicular to the plane of the loop (assumed, as is typical in such problems, so $\backslash cos\backslash theta\; =\; 1$), the change in flux is due to the change in the magnetic field's magnitude:

$\backslash phi\; =\; AB$

Therefore, the rate of change of flux is:

$$e\; =\; \backslash left|\; \backslash frac\{d(AB)\}\{dt\}\; \backslash right|\; =\; A\; \backslash left|\; \backslash frac\{dB\}\{dt\}\; \backslash right|$$

Substituting the values:

$e\; =\; (16\; \backslash times\; 10^\{-4\}\; \backslash text\{\; m\}^2)\; \backslash times\; (0.02\; \backslash text\{\; T/s\})\; =\; 0.32\; \backslash times\; 10^\{-4\}\; \backslash text\{\; V\}$

The resistance of the loop is given as $R\; =\; 1.6\; \backslash Omega$.

The induced current ($i$) in the loop can be calculated using Ohm's law:

$i\; =\; \backslash frac\{e\}\{R\}$

$i\; =\; \backslash frac\{0.32\; \backslash times\; 10^\{-4\}\; \backslash text\{\; V\}\}\{1.6\; \backslash Omega\}\; =\; 0.2\; \backslash times\; 10^\{-4\}\; \backslash text\{\; A\}\; =\; 2\; \backslash times\; 10^\{-5\}\; \backslash text\{\; A\}$

The power dissipated by the loop as heat ($P$) is given by the formula $P\; =\; i^2\; R$:

$P\; =\; (2\; \backslash times\; 10^\{-5\}\; \backslash text\{\; A\})^2\; \backslash times\; (1.6\; \backslash Omega)$

$P\; =\; (4\; \backslash times\; 10^\{-10\}\; \backslash text\{\; A\}^2)\; \backslash times\; (1.6\; \backslash Omega)\; =\; 6.4\; \backslash times\; 10^\{-10\}\; \backslash text\{\; W\}$

Source of Power: The power dissipated as heat originates from the work done by an external agent that is reducing the current in the electromagnet, thereby decreasing the magnetic field. This work done against the induced back EMF is converted into heat energy in the loop's resistance.

Solution:

We are given a square loop moving in a non-uniform and time-varying magnetic field. We need to find the induced current's magnitude and direction.

Given:

• Side of the square loop, $s\; =\; 12\; \backslash text\{\; cm\}\; =\; 0.12\; \backslash text\{\; m\}$
• Velocity of the loop, $v\; =\; 8\; \backslash text\{\; cm/s\}\; =\; 0.08\; \backslash text\{\; m/s\}$ in the positive x-direction.
• Magnetic field is in the positive z-direction.
• Gradient of the magnetic field along the negative x-direction: $\backslash frac\{dB\}\{dx\_\{neg\}\}\; =\; 10^\{-3\}\; \backslash text\{\; T\; cm\}^\{-1\}$. This means the field increases by $10^\{-3\}\; \backslash text\{\; T\}$ for every cm moved in the negative x-direction. Equivalently, it decreases by $10^\{-3\}\; \backslash text\{\; T\; cm\}^\{-1\}$ as we move in the positive x-direction.
• Rate of decrease of the magnetic field with time: $\backslash frac\{dB\}\{dt\}\; =\; -10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\}$ (negative sign indicates decrease).
• Resistance of the loop, $R\; =\; 4.50\; \backslash text\{\; m\}\backslash Omega\; =\; 4.50\; \backslash times\; 10^\{-3\}\; \backslash Omega$.

The total induced EMF ($e$) in the loop is due to two effects: the change in magnetic flux because the field is changing with time ($\backslash frac\{dB\}\{dt\}$) and the change in magnetic flux because the field is non-uniform in space and the loop is moving ($v\; \backslash times\; \backslash text\{gradient\}$).

The area of the loop is $A\; =\; s^2\; =\; (0.12\; \backslash text\{\; m\})^2\; =\; 0.0144\; \backslash text\{\; m\}^2$.

The magnetic flux through the loop is $\backslash phi\; =\; B\; A$ (assuming the field is perpendicular to the loop area).

The total rate of change of flux is given by:

$$\backslash frac\{d\backslash phi\}\{dt\}\; =\; \backslash frac\{d(BA)\}\{dt\}\; =\; A\; \backslash frac\{dB\}\{dt\}\; +\; B\; \backslash frac\{dA\}\{dt\}$$

Since the loop is rigid and moving, its area $A$ is constant, so $\backslash frac\{dA\}\{dt\}\; =\; 0$. However, the magnetic field $B$ itself is changing both with time and position. The total EMF is the sum of the EMF due to the time variation of the field and the motional EMF due to the loop's motion in the spatial gradient of the field.

The EMF due to the time-varying magnetic field is:

$e\_t\; =\; -A\; \backslash frac\{dB\}\{dt\}\; =\; -(0.0144\; \backslash text\{\; m\}^2)\; \backslash times\; (-10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\})\; =\; 1.44\; \backslash times\; 10^\{-5\}\; \backslash text\{\; V\}$

The EMF due to the motion of the loop in the spatial gradient of the magnetic field (motional EMF) needs careful consideration. The magnetic field increases as we move in the negative x-direction, meaning it decreases as we move in the positive x-direction. The gradient is $\backslash frac\{dB\}\{dx\}\; =\; -10^\{-3\}\; \backslash text\{\; T\; m\}^\{-1\}$ (since it increases in the negative x-direction, it decreases in the positive x-direction).

The motional EMF is given by $e\_m\; =\; \backslash oint\; (\backslash vec\{v\}\; \backslash times\; \backslash vec\{B\})\; \backslash cdot\; d\backslash vec\{l\}$. For a loop moving in the x-direction with a field in the z-direction that varies along x, the EMF induced across the width of the loop (in the y-direction) is relevant. The field varies along the length of the loop (in the x-direction).

Consider the two sides of the loop parallel to the y-axis. Let the loop extend from $x$ to $x+s$. The magnetic field at position $x$ is $B(x)$ and at $x+s$ is $B(x+s)$. The field changes as $B(x+s)\; \backslash approx\; B(x)\; +\; s\; \backslash frac\{dB\}\{dx\}$.

The motional EMF across the width of the loop is $e\_m\; =\; v\; \backslash times\; (\backslash text\{change\; in\; B\; across\; the\; loop\})$. The change in B across the loop's length (12 cm) due to the gradient is $s\; \backslash frac\{dB\}\{dx\}\; =\; (0.12\; \backslash text\{\; m\})\; \backslash times\; (-10^\{-3\}\; \backslash text\{\; T\; m\}^\{-1\})\; =\; -1.2\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}$.

The motional EMF is $e\_m\; =\; v\; \backslash times\; (\backslash text\{width\})\; \backslash times\; (\backslash text\{gradient\})$. However, a simpler way is to consider the flux change due to motion. The flux change rate due to motion is $\backslash frac\{d\backslash phi\_m\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}(B(x)\; A)$ where $x\; =\; vt$. So $\backslash frac\{d\backslash phi\_m\}\{dt\}\; =\; A\; \backslash frac\{dB\}\{dx\}\; \backslash frac\{dx\}\{dt\}\; =\; A\; \backslash frac\{dB\}\{dx\}\; v$.

Using the gradient $\backslash frac\{dB\}\{dx\}\; =\; -10^\{-3\}\; \backslash text\{\; T\; m\}^\{-1\}$ and $v\; =\; 0.08\; \backslash text\{\; m/s\}$, $A\; =\; 0.0144\; \backslash text\{\; m\}^2$:

$e\_m\; =\; A\; v\; \backslash frac\{dB\}\{dx\}\; =\; (0.0144\; \backslash text\{\; m\}^2)\; \backslash times\; (0.08\; \backslash text\{\; m/s\})\; \backslash times\; (-10^\{-3\}\; \backslash text\{\; T\; m\}^\{-1\})\; =\; -1.152\; \backslash times\; 10^\{-6\}\; \backslash text\{\; V\}$

The total induced EMF is the sum of these two effects. The question implies the magnetic field is in the +z direction. The loop moves in the +x direction. The gradient is along -x direction. So B increases as x decreases. As the loop moves in +x, the B field it experiences decreases.

Let's re-evaluate the gradient effect. The field increases by $10^\{-3\}\; \backslash text\{\; T\; cm\}^\{-1\}$ as we move in the negative x-direction. This means $\backslash frac\{dB\}\{dx\}\; =\; -10^\{-3\}\; \backslash text\{\; T\; cm\}^\{-1\}\; =\; -10^\{-1\}\; \backslash text\{\; T\; m\}^\{-1\}$.

The motional EMF is $e\_m\; =\; \backslash oint\; (\backslash vec\{v\}\; \backslash times\; \backslash vec\{B\})\; \backslash cdot\; d\backslash vec\{l\}$. With $\backslash vec\{v\}\; =\; v\; \backslash hat\{i\}$ and $\backslash vec\{B\}\; =\; B(x,t)\; \backslash hat\{k\}$, $\backslash vec\{v\}\; \backslash times\; \backslash vec\{B\}\; =\; v\; \backslash hat\{i\}\; \backslash times\; B(x,t)\; \backslash hat\{k\}\; =\; -v\; B(x,t)\; \backslash hat\{j\}$. The sides parallel to the y-axis have length $s$ and are oriented along $\backslash pm\; \backslash hat\{j\}$. Let's consider the EMF induced across the width $s$ in the y-direction. The field $B$ varies with $x$. The EMF induced across the width $s$ is $e\_m\; =\; s\; (\backslash vec\{v\}\; \backslash times\; \backslash vec\{B\})\_y\; =\; s\; (-v\; B\_z)\; =\; -s\; v\; B(x,t)$. This is not correct. The EMF is induced along the length element $d\backslash vec\{l\}$. The sides parallel to the y-axis are $d\backslash vec\{l\}\; =\; \backslash pm\; dy\; \backslash hat\{j\}$. The sides parallel to the x-axis are $d\backslash vec\{l\}\; =\; dx\; \backslash hat\{i\}$.

The EMF induced across the sides parallel to the y-axis (length $s$) is $e\_\{y\}\; =\; \backslash int\_\{0\}^\{s\}\; (\backslash vec\{v\}\; \backslash times\; \backslash vec\{B\})\; \backslash cdot\; (dy\; \backslash hat\{j\})\; =\; \backslash int\_\{0\}^\{s\}\; (-v\; B\_z)\; dy\; =\; -v\; B\_z\; s$. This is the same for both sides, but at different positions $x$ and $x+s$. The net EMF is the difference.

Let's use the flux change method: $\backslash frac\{d\backslash phi\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}\; (B(x(t),\; t)\; \backslash cdot\; A)$. Here $x(t)\; =\; v\; t$ (assuming $x=0$ at $t=0$). The field $B$ depends on $x$ and $t$. The gradient is $\backslash frac\{\backslash partial\; B\}\{\backslash partial\; x\}\; =\; -10^\{-1\}\; \backslash text\{\; T\; m\}^\{-1\}$ and $\backslash frac\{\backslash partial\; B\}\{\backslash partial\; t\}\; =\; -10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\}$.

The rate of change of flux is $\backslash frac\{d\backslash phi\}\{dt\}\; =\; A\; \backslash frac\{dB\}\{dt\}$, where $\backslash frac\{dB\}\{dt\}\; =\; \backslash frac\{\backslash partial\; B\}\{\backslash partial\; t\}\; +\; \backslash frac\{\backslash partial\; B\}\{\backslash partial\; x\}\; \backslash frac\{dx\}\{dt\}\; =\; \backslash frac\{\backslash partial\; B\}\{\backslash partial\; t\}\; +\; \backslash frac\{\backslash partial\; B\}\{\backslash partial\; x\}\; v$.

$\backslash frac\{dB\}\{dt\}\; =\; (-10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\})\; +\; (-10^\{-1\}\; \backslash text\{\; T\; m\}^\{-1\})\; \backslash times\; (0.08\; \backslash text\{\; m\; s\}^\{-1\})$

$\backslash frac\{dB\}\{dt\}\; =\; -10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\}\; -\; 0.08\; \backslash times\; 10^\{-1\}\; \backslash text\{\; T\; s\}^\{-1\}\; =\; -10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\}\; -\; 8\; \backslash times\; 10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\}$

$\backslash frac\{dB\}\{dt\}\; =\; -9\; \backslash times\; 10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\}$

The magnitude of the induced EMF is:

$|e|\; =\; \backslash left|\; A\; \backslash frac\{dB\}\{dt\}\; \backslash right|\; =\; (0.0144\; \backslash text\{\; m\}^2)\; \backslash times\; |-9\; \backslash times\; 10^\{-3\}\; \backslash text\{\; T\; s\}^\{-1\}|$

$|e|\; =\; 0.0144\; \backslash times\; 9\; \backslash times\; 10^\{-3\}\; \backslash text\{\; V\}\; =\; 129.6\; \backslash times\; 10^\{-5\}\; \backslash text\{\; V\}\; =\; 1.296\; \backslash times\; 10^\{-4\}\; \backslash text\{\; V\}$

The induced current ($i$) is:

$i\; =\; \backslash frac\{|e|\}\{R\}\; =\; \backslash frac\{1.296\; \backslash times\; 10^\{-4\}\; \backslash text\{\; V\}\}\{4.50\; \backslash times\; 10^\{-3\}\; \backslash Omega\}$

$i\; =\; \backslash frac\{1.296\}\{4.50\}\; \backslash times\; 10^\{-1\}\; \backslash text\{\; A\}\; \backslash approx\; 0.288\; \backslash times\; 10^\{-1\}\; \backslash text\{\; A\}\; =\; 2.88\; \backslash times\; 10^\{-2\}\; \backslash text\{\; A\}$

Direction of Induced Current: The net rate of change of flux is negative ($\backslash frac\{dB\}\{dt\}$ is negative). This means the magnetic flux in the +z direction is decreasing. According to Lenz's law, the induced current will create a magnetic field that opposes this decrease, i.e., it will try to create a magnetic field in the +z direction. Using the right-hand rule, a current flowing counter-clockwise when viewed from the positive z-axis will produce a magnetic field in the +z direction. Therefore, the induced current flows counter-clockwise in the loop.

Magnitude of Induced Current: $2.88\; \backslash times\; 10^\{-2\}\; \backslash text\{\; A\}$ or $28.8\; \backslash text\{\; mA\}$.