Electromagnetic InductionClass 12 Physics NCERT Solutions

The direction of the induced current is determined by Lenz's law, which states that the induced current will flow in a direction that opposes the change in magnetic flux that produced it.

(a) The South-pole of the magnet is moving towards the coil. This increases the magnetic flux directed into the coil from the right. To oppose this increase, the induced current will create a magnetic field directed to the left. This means the right face of the coil will act as a South-pole. Using the right-hand grip rule, the current will flow in the clockwise direction as seen from the magnet's side. The direction of the current is along qrpq.

(b) In coil pqr, the South-pole of the magnet is moving towards it. This increases the magnetic flux to the right. To oppose this, the induced current will create a magnetic field to the left, making the left face a South-pole. The current direction is clockwise, along prqp. In coil xyz, the North-pole is moving away from it. This decreases the magnetic flux to the left. To oppose this decrease, the induced current will create a magnetic field to the left, making the right face a South-pole to attract the receding North-pole. The current direction is clockwise, along yzxy.

(c) When the tapping key is just released, the current in the right-hand coil, which was flowing clockwise, decreases to zero. This causes the magnetic field pointing into the page through the left-hand coil to decrease. To oppose this decrease, the induced current in the left-hand coil will create a magnetic field directed into the page. Using the right-hand grip rule, the current will flow in the clockwise direction, along yzxy.

(d) A steady current in the straight wire produces a magnetic field directed into the plane of the loop (using the right-hand thumb rule). When the rheostat setting is changed to decrease the current, the magnetic flux into the loop decreases. To oppose this decrease, the induced current in the loop will create a magnetic field into the page. Therefore, the induced current will flow in the clockwise direction.

(e) When the tapping key is just closed, the current in the right-hand coil increases from zero, producing an increasing magnetic field pointing into the page through the left-hand coil. To oppose this increase in flux, the induced current in the left-hand coil will create a magnetic field directed out of the page. Using the right-hand grip rule, the current will flow in the counter-clockwise direction, along zyxz.

(f) The current in the straight wire is constant and directed from right to left. This creates a magnetic field directed out of the plane of the loop. As the loop moves away from the wire, the strength of the magnetic field decreases, so the magnetic flux out of the plane of the loop decreases. To oppose this decrease, the induced current will create a magnetic field out of the page. Therefore, the induced current will flow in the counter-clockwise direction.

According to Lenz's law, the induced current opposes the change in magnetic flux.

(a) A wire of irregular shape turning into a circular shape: The magnetic field is directed into the plane of the paper. When the irregular wire turns into a circular shape, its area increases. For a given perimeter, a circle encloses the maximum area. This increase in area causes an increase in the magnetic flux linked with the loop (since $\Phi_B = B \cdot A$). To oppose this increase in flux directed into the page, the induced current must produce a magnetic field directed out of the page. According to the right-hand rule, the current will flow in the counter-clockwise direction, i.e., along the path adcba.

(b) A circular loop being deformed into a narrow straight wire: The magnetic field is directed out of the plane of the paper. When the circular loop is deformed into a narrow straight wire, its area decreases. This decrease in area causes a decrease in the magnetic flux linked with the loop. To oppose this decrease in flux directed out of the page, the induced current must produce a magnetic field also directed out of the page. According to the right-hand rule, the current will flow in the counter-clockwise direction, i.e., along the path a'd'c'b'a'.

Given: Number of turns per unit length of the solenoid, $n = 15 \text{ turns/cm} = 1500 \text{ turns/m}$ Area of the small loop, $A = 2.0 \text{ cm}^2 = 2.0 \times 10^{-4} \text{ m}^2$ Initial current, $I_1 = 2.0 \text{ A}$ Final current, $I_2 = 4.0 \text{ A}$ Time interval, $\Delta t = 0.1 \text{ s}$

To Find: The induced emf in the loop, $\varepsilon$.

Formula: The magnetic field inside a long solenoid is given by $B = \mu_0 n I$. The magnetic flux through the small loop is $\Phi_B = B A = (\mu_0 n I) A$, since the loop is placed normal to the axis. According to Faraday's law of induction, the induced emf is: $\varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(\mu_0 n A I) = -\mu_0 n A \frac{dI}{dt}$ For a steady change in current, $\frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{I_2 - I_1}{\Delta t}$. So, the magnitude of the induced emf is $|\varepsilon| = \mu_0 n A \left|\frac{I_2 - I_1}{\Delta t}\right|$.

Calculation: Rate of change of current, $\frac{dI}{dt} = \frac{4.0 \text{ A} - 2.0 \text{ A}}{0.1 \text{ s}} = \frac{2.0 \text{ A}}{0.1 \text{ s}} = 20 \text{ A/s}$. We know that $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.

Substituting the values into the formula for the magnitude of emf: $|\varepsilon| = (4\pi \times 10^{-7} \text{ T m/A}) \times (1500 \text{ m}^{-1}) \times (2.0 \times 10^{-4} \text{ m}^2) \times (20 \text{ A/s})$ $|\varepsilon| = 4 \times 3.14 \times 10^{-7} \times 1500 \times 2.0 \times 10^{-4} \times 20$ $|\varepsilon| = (12.56 \times 10^{-7}) \times (60000 \times 10^{-4})$ $|\varepsilon| = 12.56 \times 6 \times 10^{-7} = 75.36 \times 10^{-7} \text{ V}$ $|\varepsilon| = 7.536 \times 10^{-6} \text{ V} = 7.54 \ \mu\text{V}$

Final Answer: The induced emf in the loop while the current is changing is $7.54 \times 10^{-6} \text{ V}$ or $7.54 \ \mu\text{V}$. Note: The provided solution in the source has a calculation error. Recalculating: $4\pi \times 1500 \times 2 \times 10^{-4} \times 20 \times 10^{-7} = 4\pi \times 6 \times 10^{-7} = 24\pi \times 10^{-7} \approx 75.4 \times 10^{-7} = 7.54 \times 10^{-6}$ V. The calculation above is correct.

Given: Longer side of the loop, $L = 8 \text{ cm} = 0.08 \text{ m}$ Shorter side of the loop, $b = 2 \text{ cm} = 0.02 \text{ m}$ Magnitude of the magnetic field, $B = 0.3 \text{ T}$ Velocity of the loop, $v = 1 \text{ cm/s} = 0.01 \text{ m/s}$

To Find: The emf developed ($\varepsilon$) and the duration for which it lasts ($\Delta t$) in two cases.

Formula: The motional emf induced in a conductor of length $l$ moving with velocity $v$ perpendicular to a magnetic field $B$ is given by: $\varepsilon = B l v$

The induced voltage lasts as long as the loop is exiting the field. The duration is the time taken to travel a distance equal to the side of the loop that is parallel to the direction of motion. $\Delta t = \frac{\text{distance}}{\text{velocity}}$

Case (a): Velocity normal to the longer side In this case, the length of the conductor moving perpendicular to the field is the longer side, so $l = L = 0.08 \text{ m}$.

Calculation (emf): $\varepsilon = B L v = (0.3 \text{ T}) \times (0.08 \text{ m}) \times (0.01 \text{ m/s})$ $\varepsilon = 0.00024 \text{ V} = 2.4 \times 10^{-4} \text{ V}$

Calculation (duration): The loop moves a distance equal to its shorter side to exit the field. So, distance $= b = 0.02 \text{ m}$. $\Delta t = \frac{b}{v} = \frac{0.02 \text{ m}}{0.01 \text{ m/s}} = 2 \text{ s}$

Case (b): Velocity normal to the shorter side In this case, the length of the conductor moving perpendicular to the field is the shorter side, so $l = b = 0.02 \text{ m}$.

Calculation (emf): $\varepsilon = B b v = (0.3 \text{ T}) \times (0.02 \text{ m}) \times (0.01 \text{ m/s})$ $\varepsilon = 0.00006 \text{ V} = 6.0 \times 10^{-5} \text{ V}$

Calculation (duration): The loop moves a distance equal to its longer side to exit the field. So, distance $= L = 0.08 \text{ m}$. $\Delta t = \frac{L}{v} = \frac{0.08 \text{ m}}{0.01 \text{ m/s}} = 8 \text{ s}$

Final Answer: (a) When moving normal to the longer side: The induced emf is $2.4 \times 10^{-4} \text{ V}$ and it lasts for $2 \text{ s}$. (b) When moving normal to the shorter side: The induced emf is $6.0 \times 10^{-5} \text{ V}$ and it lasts for $8 \text{ s}$.

Given: Length of the metallic rod, $L = 1.0 \text{ m}$ Angular frequency, $\omega = 400 \text{ rad/s}$ Magnetic field, $B = 0.5 \text{ T}$

To Find: The emf developed between the centre and the ring, $\varepsilon$.

Formula: Consider a small element of length $dr$ of the rod at a distance $r$ from the centre. The linear velocity of this element is $v = r\omega$. This element moves perpendicular to the magnetic field. The small emf induced across this element is $d\varepsilon = B v dr = B (r\omega) dr$. The total emf developed between the centre (r=0) and the end of the rod (r=L) is the integral of $d\varepsilon$ from 0 to L. $\varepsilon = \int_0^L d\varepsilon = \int_0^L B\omega r \, dr$ $\varepsilon = B\omega \int_0^L r \, dr = B\omega \left[\frac{r^2}{2}\right]_0^L$ $\varepsilon = \frac{1}{2} B \omega L^2$

Calculation: Substituting the given values into the formula: $\varepsilon = \frac{1}{2} \times (0.5 \text{ T}) \times (400 \text{ rad/s}) \times (1.0 \text{ m})^2$ $\varepsilon = \frac{1}{2} \times 0.5 \times 400 \times 1$ $\varepsilon = 0.25 \times 400$ $\varepsilon = 100 \text{ V}$

Final Answer: The emf developed between the centre and the ring is $100 \text{ V}$.

Given: Length of the wire, $l = 10 \text{ m}$ Speed of the wire, $v = 5.0 \text{ m/s}$ (falling downwards) Horizontal component of Earth's magnetic field, $B_H = 0.30 \times 10^{-4} \text{ Wb m}^{-2}$ (directed from South to North)

To Find: (a) Instantaneous induced emf, $\varepsilon$. (b) Direction of the emf. (c) The end of the wire with higher electrical potential.

Formula: The wire is falling at right angles to $B_H$. The motional emf is given by: $\varepsilon = B_H l v$

Calculation: $\varepsilon = (0.30 \times 10^{-4} \text{ T}) \times (10 \text{ m}) \times (5.0 \text{ m/s})$ $\varepsilon = 1.5 \times 10^{-3} \text{ V} = 1.5 \text{ mV}$

(b) Direction of the emf and (c) End at higher potential

Method: We can determine the direction using the Lorentz force on the free electrons in the wire, $\vec{F} = q(\vec{v} \times \vec{B})$. Here, $q$ is negative for electrons. Alternatively, we can use Fleming's Right-Hand Rule for a positive charge carrier.

Using the vector cross product for a positive charge, the direction of the force $\vec{F}$ is given by $\vec{v} \times \vec{B}_H$. Pointing the fingers of the right hand downwards (for $\vec{v}$) and curling them towards the North (for $\vec{B}_H$), the thumb points towards the East. This means that positive charges in the wire are pushed towards the eastern end.

Therefore, the eastern end of the wire accumulates positive charge and becomes at a higher electrical potential, while the western end becomes at a lower potential. The direction of the induced emf (and conventional current) is from the lower potential end to the higher potential end, which is from West to East.

Final Answer: (a) The instantaneous value of the induced emf is $1.5 \times 10^{-3} \text{ V}$. (b) The direction of the emf is from the western end to the eastern end of the wire. (c) The eastern end of the wire is at the higher electrical potential.

Given: Initial current, $I_1 = 5.0 \text{ A}$ Final current, $I_2 = 0.0 \text{ A}$ Time interval, $\Delta t = 0.1 \text{ s}$ Average induced emf, $\varepsilon = 200 \text{ V}$

To Find: The self-inductance of the circuit, $L$.

Formula: The relationship between self-inductance, induced emf, and the rate of change of current is given by Faraday's law for an inductor: $\varepsilon = -L \frac{dI}{dt}$ For an average emf and a steady change in current, we can write the magnitude of the emf as: $|\varepsilon| = L \left|\frac{\Delta I}{\Delta t}\right| = L \left|\frac{I_2 - I_1}{\Delta t}\right|$ Rearranging for $L$: $L = \frac{|\varepsilon|}{\left|\frac{\Delta I}{\Delta t}\right|}$

Calculation: First, calculate the magnitude of the rate of change of current: $\left|\frac{\Delta I}{\Delta t}\right| = \left|\frac{0.0 \text{ A} - 5.0 \text{ A}}{0.1 \text{ s}}\right| = \left|\frac{-5.0 \text{ A}}{0.1 \text{ s}}\right| = 50 \text{ A/s}$

Now, calculate the self-inductance $L$: $L = \frac{200 \text{ V}}{50 \text{ A/s}} = 4.0 \text{ H}$

Final Answer: The estimated self-inductance of the circuit is $4.0 \text{ H}$.

Given: Mutual inductance, $M = 1.5 \text{ H}$ Initial current in the first coil, $I_1 = 0 \text{ A}$ Final current in the first coil, $I_2 = 20 \text{ A}$ Time interval, $\Delta t = 0.5 \text{ s}$ (This information is not needed to find the change in flux linkage, but would be needed for the induced emf).

To Find: The change of flux linkage with the other coil, $\Delta \Phi_{\text{linkage}}$.

Formula: The magnetic flux linkage ($N\Phi_B$) with the second coil is directly proportional to the current ($I$) in the first coil. The constant of proportionality is the mutual inductance ($M$). $\Phi_{\text{linkage}} = N\Phi_B = M I$

The change in flux linkage is therefore given by: $\Delta \Phi_{\text{linkage}} = M \Delta I = M (I_2 - I_1)$

Calculation: First, calculate the change in current: $\Delta I = I_2 - I_1 = 20 \text{ A} - 0 \text{ A} = 20 \text{ A}$

Now, calculate the change in flux linkage: $\Delta \Phi_{\text{linkage}} = (1.5 \text{ H}) \times (20 \text{ A})$ $\Delta \Phi_{\text{linkage}} = 30 \text{ Wb}$ (Note: The unit of flux linkage is Weber, since Henry is Volt-second/Ampere, and Volt-second is Weber).

Final Answer: The change of flux linkage with the other coil is $30 \text{ Wb}$.