Q.A long solenoid with $15$ turns per cm has a small loop of area $2.0\ \text{cm}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0\ \text{A}$ to $4.0\ \text{A}$ in $0.1\ \text{s}$, what is the induced emf in the loop while the current is changing?

Showing the 12 most recent of 28 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.When an inductor is connected to a 200 V dc supply, the current through it is 5 A and when the same inductor is connected to a 200 V ac supply of angular frequency $300~\mathrm{rads}^{-1}$, the current through it is 4 A. The inductance of the inductor is (A) $100~\mathrm{mH}$ (B) $200~\mathrm{mH}$ (C) $50~\mathrm{mH}$ (D) $75~\mathrm{mH}$

   ›Reveal solutionSolution

   The dc case gives the coil’s resistance; the ac case gives its impedance; the difference yields the inductive reactance and hence the inductance. The answer is 100 mH.

   The key idea is that an inductor is not a pure inductor in practice — it has some internal resistance due to the wire it’s wound from. When you connect it to a dc supply, only that resistance matters because the inductive reactance is zero for steady current. When you connect it to an ac supply, both the resistance and the inductive reactance oppose the current, so the impedance is larger and the current smaller. Comparing the two cases lets you isolate the inductance.

   Let’s work it through.

   1. Find the resistance from the dc case. For a dc supply, the inductor behaves as a pure resistor $R$ (the coil’s internal resistance). Ohm’s law gives

   $R = \frac{V_{\text{dc}}}{I_{\text{dc}}} = \frac{200\ \text{V}}{5\ \text{A}} = 40\ \Omega.$

   2. Find the impedance from the ac case. For the ac supply (rms values), the impedance $Z$ of the series $RL$ combination is

   $Z = \frac{V_{\text{ac}}}{I_{\text{ac}}} = \frac{200\ \text{V}}{4\ \text{A}} = 50\ \Omega.$

   3. Relate impedance to resistance and inductive reactance. For an $RL$ series circuit,

   $Z = \sqrt{R^2 + X_L^2}, \quad \text{where } X_L = \omega L.$

   Substitute the known values:

   $50 = \sqrt{40^2 + (\omega L)^2}.$

   4. Solve for $L$. Square both sides:

   $2500 = 1600 + (\omega L)^2 \quad\Rightarrow\quad (\omega L)^2 = 900.$

   So $\omega L = 30\ \Omega$ (taking the positive root, as reactance is positive).

   Given $\omega = 300\ \text{rad/s}$,

   $L = \frac{30}{300} = 0.1\ \text{H} = 100\ \text{mH}.$

   Watch out

   A common mistake is to treat the inductor as ideal (zero resistance) and directly use $X_L = V/I$ from the ac case. That gives $L = 200/ (4 \times 300) \approx 167\ \text{mH}$, which is not among the options — a clue that resistance must be present.

   Tip

   Whenever a problem gives both dc and ac data for the same coil, the dc data always gives the resistance. The ac data then gives the impedance, and the difference is the inductive reactance.

   ✓Final answer

   The inductance of the inductor is $\boxed{100\ \text{mH}}$, which corresponds to option (A).

2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If the length of a linear antenna is increased by $60\%$ and the wavelength of the signal is decreased by $20\%$, then the percentage increase in the effective power radiated by the antenna is (A) $50$ (B) $250$ (C) $300$ (D) $150$

   ›Reveal solutionSolution

   The effective power radiated by a linear antenna is proportional to $\left(\frac{L}{\lambda}\right)^2$. Increasing $L$ by $60\%$ and decreasing $\lambda$ by $20\%$ multiplies $\frac{L}{\lambda}$ by $2$, so power increases by $300\%$. The answer is (C).

   The key idea here is that for a linear antenna (like a half-wave dipole or a simple straight wire), the effective power radiated is not simply proportional to the length or the wavelength separately — it depends on the ratio of the antenna length to the wavelength. This comes from the fact that the radiation efficiency and the radiation resistance both scale with $(L/\lambda)^2$ when the antenna is much shorter than the wavelength (a "short dipole" approximation). Even for a resonant antenna, the power radiated for a given current is proportional to the radiation resistance, which scales as $(L/\lambda)^2$ for a short dipole. The problem implicitly uses this proportionality.

   Let’s work through it step by step.

   1. Set up the proportionality. The effective power radiated $P$ is proportional to $\left(\frac{L}{\lambda}\right)^2$, where $L$ is the antenna length and $\lambda$ is the wavelength of the signal.

   $P \propto \left(\frac{L}{\lambda}\right)^2$

   2. Express the changes. The length is increased by $60\%$, so the new length $L'$ is:

   $L' = L + 0.60L = 1.60L$

   The wavelength is decreased by $20\%$, so the new wavelength $\lambda'$ is:

   $\lambda' = \lambda - 0.20\lambda = 0.80\lambda$

   3. Find the new ratio. The new ratio $\frac{L'}{\lambda'}$ becomes:

   $\frac{L'}{\lambda'} = \frac{1.60L}{0.80\lambda} = \frac{1.60}{0.80} \cdot \frac{L}{\lambda} = 2 \cdot \frac{L}{\lambda}$

   So the ratio doubles.

   4. Find the new power. Since $P \propto (L/\lambda)^2$, the new power $P'$ is:

   $P' \propto \left(2 \cdot \frac{L}{\lambda}\right)^2 = 4 \left(\frac{L}{\lambda}\right)^2 \propto 4P$

   That means $P' = 4P$, i.e., the power becomes four times the original.

   5. Calculate the percentage increase. Percentage increase = $\frac{P' - P}{P} \times 100\% = \frac{4P - P}{P} \times 100\% = 3 \times 100\% = 300\%$.

   Watch out

   A common mistake is to treat the power as proportional to $L$ alone or to $1/\lambda$ alone. If you only increased $L$ by $60\%$, power would increase by only $156\%$ (since $1.6^2 = 2.56$). If you only decreased $\lambda$ by $20\%$, power would increase by $56.25\%$ (since $1/0.8^2 = 1.5625$). But the combined effect multiplies, not adds — the ratio doubles, so power quadruples.

   Tip

   Whenever you see "percentage increase" in a problem involving a ratio squared, first find the factor by which the ratio changes. Here, $L$ multiplies by $1.6$ and $\lambda$ multiplies by $0.8$, so the ratio multiplies by $1.6/0.8 = 2$. Square that to get $4$, meaning a $300\%$ increase.

   ✓Final answer

   The percentage increase in the effective power radiated is $300\%$, so the correct option is (C).

3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.A circular coil of radius 10 cm is carrying a current. From the centre of the coil, the distance of a point on its axis where $\left(\frac{dB}{dx}\right)$ becomes maximum is ($\frac{dB}{dx}$ is the change in magnetic field with distance x from the centre of the coil) (A) 40 cm (B) 20 cm (C) 10 cm (D) 5 cm

   ›Reveal solutionSolution

   The magnetic field on the axis of a circular coil varies with distance $x$, and its derivative $\frac{dB}{dx}$ is maximised where the second derivative of $B$ vanishes. For a coil of radius $R$, this occurs at $x = \frac{R}{2}$. With $R = 10$ cm, the answer is $5$ cm.

   The magnetic field on the axis of a circular coil of radius $R$, carrying a current $I$, at a distance $x$ from its centre is given by

   $B(x) = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}.$

   This is a standard result derived from the Biot–Savart law — the field is symmetric about the centre and falls off as you move away. The question asks where the rate of change of this field, $\frac{dB}{dx}$, is itself maximum. That means we are looking for the point where the slope of $B(x)$ is steepest.

   Why not just differentiate $B$ and find where that derivative is largest? Because $\frac{dB}{dx}$ is itself a function of $x$, and its maximum occurs where its own derivative (the second derivative of $B$) is zero and changes sign. So the plan is:

   1. Write $B(x)$.
   2. Compute $\frac{dB}{dx}$.
   3. Differentiate again to get $\frac{d^2B}{dx^2}$.
   4. Set $\frac{d^2B}{dx^2} = 0$ and solve for $x$.

   Let’s do it step by step.

   1. Write $B(x)$ and factor constants. Let $k = \frac{\mu_0 I R^2}{2}$, a positive constant. Then

   $B(x) = k (R^2 + x^2)^{-3/2}.$

   2. First derivative. Using the chain rule:

   $\frac{dB}{dx} = k \cdot \left(-\frac{3}{2}\right) (R^2 + x^2)^{-5/2} \cdot 2x = -3k x (R^2 + x^2)^{-5/2}.$

   The negative sign just tells us the field decreases as $x$ increases — we care about the magnitude of the slope, but the location of the extremum of $\frac{dB}{dx}$ is the same whether we consider the signed derivative or its absolute value.

   3. Second derivative. Differentiate $\frac{dB}{dx}$ using the product rule. Let $u = -3k x$ and $v = (R^2 + x^2)^{-5/2}$. Then

   $\frac{d^2B}{dx^2} = u' v + u v'.$

   We have $u' = -3k$, and

   $v' = -\frac{5}{2} (R^2 + x^2)^{-7/2} \cdot 2x = -5x (R^2 + x^2)^{-7/2}.$

   So

   $\frac{d^2B}{dx^2} = (-3k)(R^2 + x^2)^{-5/2} + (-3k x)\left[-5x (R^2 + x^2)^{-7/2}\right].$

   Simplify the second term: $(-3k x)(-5x) = 15k x^2$, so

   $\frac{d^2B}{dx^2} = -3k (R^2 + x^2)^{-5/2} + 15k x^2 (R^2 + x^2)^{-7/2}.$

   4. Factor to set to zero. Factor out $3k (R^2 + x^2)^{-7/2}$ (which is never zero):

   $\frac{d^2B}{dx^2} = 3k (R^2 + x^2)^{-7/2} \left[ - (R^2 + x^2) + 5x^2 \right].$

   The bracket simplifies to $-R^2 - x^2 + 5x^2 = -R^2 + 4x^2$.

   Setting $\frac{d^2B}{dx^2} = 0$ gives

   $-R^2 + 4x^2 = 0 \quad \Rightarrow \quad 4x^2 = R^2 \quad \Rightarrow \quad x = \frac{R}{2}.$

   (We take the positive root since $x$ is distance from the centre.)

   Tip

   The result $x = R/2$ is independent of the current $I$ and the constant $\mu_0$ — it depends only on the geometry. For any circular coil, the steepest slope of $B(x)$ occurs at half the radius.

   5. Plug in the given radius. Here $R = 10$ cm, so

   $x = \frac{10}{2} = 5 \text{ cm}.$

   Watch out

   A common mistake is to set $\frac{dB}{dx} = 0$ instead of $\frac{d^2B}{dx^2} = 0$. That would give $x = 0$ (the centre), where the field is maximum but its rate of change is zero — not the steepest slope.

   ✓Final answer

   The distance is $5$ cm, which corresponds to option (D).

4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.In the given figure, the directions of induced current in a square loop as it enters and leaves a uniform magnetic field respectively are (A) Counter clockwise, Clockwise (B) Clockwise, Counter clockwise (C) Counter clockwise, Counter clockwise (D) Clockwise, Clockwise

   ›Reveal solutionSolution

   The key idea is Lenz’s law: induced current opposes the change in magnetic flux. As the loop enters the field, flux increases, so induced current creates opposing field (out of page) → counterclockwise. As it leaves, flux decreases, so induced current tries to sustain field (into page) → clockwise. Final answer: (A).

   Concept & Intuition

   This problem tests Lenz’s law, which tells us the direction of induced current: it always flows so that its own magnetic field opposes the change in the external magnetic flux through the loop.

   Imagine the uniform magnetic field is directed into the page (a common convention). When the square loop moves into the field, more of its area is inside the field, so the flux (into page) increases. The induced current must create a field out of the page to oppose that increase. Using the right-hand rule, a counterclockwise current produces a field out of the page.

   When the loop leaves the field, the flux (into page) decreases. Now the induced current tries to sustain the field into the page, so it must produce a field into the page — which requires a clockwise current.

   Step-by-step reasoning

   1. Identify the external field direction

      The problem doesn’t explicitly state the field direction, but by standard convention in such diagrams, the uniform magnetic field is perpendicular to the plane of the loop and directed into the page (often shown by crosses). We adopt that.

   2. Analyze the “entering” phase

      • As the loop moves into the field, the area inside the field increases → magnetic flux (into page) increases.
      • Lenz’s law: induced current opposes this increase → its own field must point out of the page.
      • Right-hand rule: curl fingers in direction of current; thumb gives field direction. For field out of page, current must be counterclockwise.
      • So first answer: counterclockwise.

   3. Analyze the “leaving” phase

      • As the loop moves out, the area inside the field decreases → flux (into page) decreases.
      • Lenz’s law: induced current opposes this decrease → its own field must point into the page (to replace the lost flux).
      • Right-hand rule: for field into page, current must be clockwise.
      • So second answer: clockwise.

   4. Match with options

      The pair (counterclockwise, clockwise) corresponds to option (A).

   Watch out

   A common mistake is to think the induced current always opposes the motion of the loop. It actually opposes the change in flux, not the motion itself. Here, motion into the field increases flux, so the induced field opposes that increase — not necessarily the motion.

   Tip

   A quick memory aid: “Entering → opposing increase → opposite field → counterclockwise; Leaving → opposing decrease → same field → clockwise.” If the external field were out of the page, the directions would reverse, but the logic is identical.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The magnetic flux ($\phi$ in weber) linked with a coil varies with time (t in second) as per the equation $\phi = 2.5t^2 + 5t + 7$. If the induced electric power in the coil at a time 3 s is 2 W, then the induced electric power in the coil at a time 5 s is (A) 5.5 W (B) 3.5 W (C) 2.5 W (D) 4.5 W

   ›Reveal solutionSolution

   The induced emf is the negative rate of change of flux, and power is $P = \mathcal{E}^2 / R$. Using the given flux equation, we find the resistance from the 3 s data, then compute power at 5 s. The result is 4.5 W, option (D).

   Concept & Intuition

   Faraday’s law tells us that a changing magnetic flux induces an electromotive force (emf) in a coil: $\mathcal{E} = -\frac{d\phi}{dt}$. The induced electric power dissipated in the coil (assuming it’s purely resistive) is $P = \frac{\mathcal{E}^2}{R}$, where $R$ is the coil’s resistance. Since the flux is given as a quadratic function of time, the emf will be linear in time. Once we know the resistance from the power at one instant, we can find the power at any other instant.

   Step-by-step solution

   1. Find the induced emf as a function of time. Given $\phi = 2.5t^2 + 5t + 7$, differentiate with respect to $t$:

   $\mathcal{E}(t) = -\frac{d\phi}{dt} = -(5t + 5) = -5(t+1)$

   The magnitude of the emf is $|\mathcal{E}(t)| = 5(t+1)$ (the sign only indicates direction; power depends on magnitude squared).

   2. Use the power at $t = 3$ s to find the coil’s resistance $R$. At $t = 3$ s:

   $|\mathcal{E}(3)| = 5(3+1) = 20 \text{ V}$

   Power is given as $P(3) = 2$ W. Using $P = \mathcal{E}^2 / R$:

   $2 = \frac{20^2}{R} \quad \Rightarrow \quad 2 = \frac{400}{R} \quad \Rightarrow \quad R = 200 \, \Omega$

   3. Compute the emf at $t = 5$ s.

   $|\mathcal{E}(5)| = 5(5+1) = 30 \text{ V}$

   4. Find the power at $t = 5$ s.

   $P(5) = \frac{\mathcal{E}(5)^2}{R} = \frac{30^2}{200} = \frac{900}{200} = 4.5 \text{ W}$

   Tip

   Notice that the constant term $+7$ in the flux disappears upon differentiation — it doesn’t affect the induced emf. Only the time-varying part matters.

   Watch out

   A common mistake is to forget that power depends on emf squared, so doubling the emf quadruples the power. Here, emf goes from 20 V to 30 V (a factor of 1.5), so power goes from 2 W to $2 \times (1.5)^2 = 4.5$ W.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The plane of a circular coil of resistance $7.5~\Omega$ is placed perpendicular to a uniform magnetic field. The flux $\phi$ (in weber) through the coil varies with time $t$ (in second) as $\phi = 2t^2 + 3t - 2$. The induced power in the coil at time $t = 3~\text{s}$ is (A) $7.5~\text{W}$ (B) $15~\text{W}$ (C) $30~\text{W}$ (D) $20~\text{W}$

   ›Reveal solutionSolution

   The induced power is found from $P = \mathcal{E}^2 / R$, where $\mathcal{E} = |d\phi/dt|$. Differentiating $\phi = 2t^2 + 3t - 2$ gives $\mathcal{E} = 4t + 3$, so at $t = 3$ s, $\mathcal{E} = 15$ V, and $P = 15^2 / 7.5 = 30$ W. The correct option is (C).

   The key idea is that induced power depends on the induced emf and the coil’s resistance. Since the coil’s plane is perpendicular to the field, the entire flux change contributes to the emf via Faraday’s law. Power is then just $P = \mathcal{E}^2 / R$, because the coil behaves like a simple resistive circuit with that emf.

   Why this works:

   Faraday’s law tells us the induced emf is the negative rate of change of magnetic flux: $\mathcal{E} = -d\phi/dt$. The magnitude matters for power (since power depends on $\mathcal{E}^2$). Once we have $\mathcal{E}$, the instantaneous power dissipated in the resistor is $P = \mathcal{E}^2 / R$. No need to worry about direction or Lenz’s law here — just the magnitude.

   Step-by-step:

   1. Find the induced emf as a function of time. Given $\phi(t) = 2t^2 + 3t - 2$, differentiate:

   $\frac{d\phi}{dt} = 4t + 3.$

   The magnitude of the induced emf is

   $|\mathcal{E}| = \left| \frac{d\phi}{dt} \right| = 4t + 3 \quad (\text{positive for } t \ge 0).$

   2. Evaluate at $t = 3$ s.

   $\mathcal{E}(3) = 4(3) + 3 = 12 + 3 = 15 \text{ V}.$

   3. Use the power formula for a resistor. The coil has resistance $R = 7.5\ \Omega$. Instantaneous power dissipated is

   $P = \frac{\mathcal{E}^2}{R} = \frac{15^2}{7.5} = \frac{225}{7.5}.$

   4. Simplify.

   $\frac{225}{7.5} = \frac{225}{15/2} = 225 \times \frac{2}{15} = 15 \times 2 = 30 \text{ W}.$

   Watch out

   A common mistake is to forget that power depends on $\mathcal{E}^2$, not $\mathcal{E}$. If you used $P = \mathcal{E} I$ and tried to find current from flux directly, you’d risk sign errors. Stick to $P = \mathcal{E}^2 / R$ — it’s simpler and foolproof.

   Tip

   Notice that the constant term $-2$ in $\phi$ disappears upon differentiation — it doesn’t affect the emf. Only the time-dependent parts matter.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A current of 4 A is passed through a square loop of side 5 cm made of a uniform manganin wire as shown in the figure. The magnetic field at the centre of the loop is (A) $\dfrac{24\sqrt{2}}{5} \times 10^{-5}$ T (B) $\dfrac{3\sqrt{2}}{5} \times 10^{-5}$ T (C) $\dfrac{6\sqrt{2}}{5} \times 10^{-5}$ T (D) Zero

   ›Reveal solutionSolution

   The net magnetic field at the centre is zero — option (D).

   Setup. The current $I = 4\,\text{A}$ enters the square loop at one point and leaves at another, splitting into two arcs (paths) of the same uniform manganin wire. Being uniform, resistance is proportional to length, so the two paths carry currents inversely proportional to their lengths:

   $i_1 R_1 = i_2 R_2 \;\Rightarrow\; i_1 L_1 = i_2 L_2.$

   Field cancellation. Each path produces a magnetic field at the centre, but the two currents circulate in opposite senses about the centre. For a wire segment, the field at the centre is proportional to $i$ times a geometric factor that scales with its length. Because $i_1 L_1 = i_2 L_2$, the two contributions are equal in magnitude and opposite in direction:

   $\vec{B}_1 + \vec{B}_2 = 0.$

   The field from the entering/leaving connecting leads (directed along the centre) adds nothing. Hence the resultant field at the centre vanishes.

   ✓Final answer

   (D) Zero

8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A coil of resistance $8~\Omega$, number of turns $250$ and area $120~\text{cm}^2$ is placed in a uniform magnetic field of $2~\text{T}$ such that the plane of the coil makes an angle of $\dfrac{\pi}{6}$ with the direction of the magnetic field. In a time of $100~\text{ms}$, the coil is rotated until its plane becomes parallel to the direction of the magnetic field. The current induced in the coil is (A) $5.25~\text{A}$ (B) $3.75~\text{A}$ (C) $2.75~\text{A}$ (D) $1.25~\text{A}$

   ›Reveal solutionSolution

   The induced current is found from Faraday’s law: the change in magnetic flux divided by the time interval gives the induced emf, and Ohm’s law gives the current. The result is 3.75 A, which corresponds to option (B).

   The key idea is that the induced emf depends on the rate of change of magnetic flux through the coil. The flux changes because the orientation of the coil relative to the magnetic field changes. We don’t need to worry about the details of the rotation — only the initial and final flux matter, because the emf is the total flux change divided by the time (assuming constant rotation rate).

   Why this works:

   Faraday’s law says the induced emf is $\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}$ for a constant rate of change. The magnetic flux through a single turn is $\Phi = B A \cos\theta$, where $\theta$ is the angle between the magnetic field and the normal to the coil’s plane. The problem gives the angle between the field and the plane of the coil, so we must convert carefully.

   ---

   1. Convert the given angle to the angle with the normal. The plane of the coil makes an angle of $\pi/6$ with the magnetic field. The normal to the plane is perpendicular to the plane, so the angle between the normal and the field is

   $\theta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}.$

   Initially, $\theta_i = \pi/3$.

   Finally, the plane is parallel to the field, so the normal is perpendicular to the field: $\theta_f = \pi/2$.

   2. Compute the initial and final magnetic flux through one turn. Area $A = 120~\text{cm}^2 = 120 \times 10^{-4}~\text{m}^2 = 0.012~\text{m}^2$. Magnetic field $B = 2~\text{T}$.

   $\Phi_i = B A \cos\theta_i = 2 \times 0.012 \times \cos(\pi/3) = 0.024 \times \frac{1}{2} = 0.012~\text{Wb}.$

   $\Phi_f = B A \cos\theta_f = 2 \times 0.012 \times \cos(\pi/2) = 0.024 \times 0 = 0~\text{Wb}.$

   3. Find the total change in flux for the entire coil. The coil has $N = 250$ turns, so the total flux linkage change is

   $\Delta \Phi_{\text{total}} = N (\Phi_f - \Phi_i) = 250 \times (0 - 0.012) = -3.0~\text{Wb}.$

   The magnitude of the change is $3.0~\text{Wb}$.

   4. Calculate the induced emf. Time interval $\Delta t = 100~\text{ms} = 0.1~\text{s}$.

   $|\mathcal{E}| = N \frac{|\Delta \Phi|}{\Delta t} = \frac{3.0}{0.1} = 30~\text{V}.$

   5. Use Ohm’s law to find the induced current. Resistance $R = 8~\Omega$.

   $I = \frac{|\mathcal{E}|}{R} = \frac{30}{8} = 3.75~\text{A}.$

   Watch out

   A common mistake is to use the angle between the plane and the field directly in the flux formula. Remember: flux uses the angle between the field and the normal to the plane. Always convert: $\theta_{\text{normal}} = 90^\circ - \theta_{\text{plane}}$.

   Tip

   Since the final flux is zero, the calculation simplifies: the induced emf depends only on the initial flux. This is a neat shortcut when the coil ends up aligned so that no field lines pass through it.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The area of cross-section of a potentiometer wire is $6 \times 10^{-7} \ \mathrm{m}^2$. The potential difference per unit length of the potentiometer wire when it is connected to a cell of negligible internal resistance and a resistor in series is $0.15 \ \mathrm{Vm}^{-1}$. If the current through potentiometer wire is $0.3 \ \mathrm{A}$, then the resistivity of the material of the potentiometer wire is (A) $4 \times 10^{-6} \ \Omega \mathrm{m}$ (B) $3 \times 10^{-7} \ \Omega \mathrm{m}$ (C) $3 \times 10^{-6} \ \Omega \mathrm{m}$ (D) $4 \times 10^{-7} \ \Omega \mathrm{m}$

   ›Reveal solutionSolution

   The resistivity is found by combining Ohm’s law in terms of electric field and current density: $\rho = \frac{E}{J} = \frac{(\text{potential gradient})}{(\text{current density})}$. Substituting the given values gives $\rho = 3 \times 10^{-7} \, \Omega \mathrm{m}$, which corresponds to option (B).

   Concept & Intuition

   The potentiometer wire is a uniform conductor carrying a steady current. The potential difference per unit length (the potential gradient) is the electric field $E$ inside the wire. The current density $J$ is the current per unit cross-sectional area. Resistivity $\rho$ relates these two via the microscopic Ohm’s law: $E = \rho J$. So instead of using $R = \rho L/A$ and then $V = IR$, we can directly compute $\rho = E/J$ — a cleaner path that avoids needing the wire’s length.

   Step-by-step solution

   1. Identify the electric field in the wire The potential difference per unit length is given as $0.15 \, \mathrm{V/m}$. This is exactly the magnitude of the uniform electric field inside the wire:

   $E = 0.15 \, \mathrm{V\,m^{-1}}.$

   2. Find the current density Current density $J$ is current per unit cross-sectional area:

   $J = \frac{I}{A} = \frac{0.3 \, \mathrm{A}}{6 \times 10^{-7} \, \mathrm{m}^2} = 5 \times 10^{5} \, \mathrm{A\,m^{-2}}.$

   3. Apply the microscopic Ohm’s law For a homogeneous conductor, $E = \rho J$. Solving for resistivity:

   $\rho = \frac{E}{J} = \frac{0.15}{5 \times 10^{5}} = 3 \times 10^{-7} \, \Omega \mathrm{m}.$

   4. Match with the options The value $3 \times 10^{-7} \, \Omega \mathrm{m}$ corresponds to option (B).

   Watch out

   A common mistake is to first compute resistance using $R = V/I$ with the potential gradient, but that requires the wire’s length, which is not given. Using $E = \rho J$ bypasses this entirely — always look for the direct relation between field and current density when length is unknown.

   Tip

   Remember: potential gradient = electric field = $\rho \times (I/A)$. This single formula is often the fastest route in potentiometer resistivity problems.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.A coil of resistance $16~\Omega$ is placed with its plane perpendicular to a uniform magnetic field whose flux ($\phi$ in $10^{-3}$ weber) changes with time (t in second) as $\phi = 5t^2 + 4t + 2$. The induced current at time $t = 6$ seconds is (A) 4 mA (B) 2.12 mA (C) 34 mA (D) 74 mA

    ›Reveal solutionSolution

    The induced current is found from Faraday’s law: induced emf = $-d\phi/dt$, then current = emf / resistance. At $t = 6$ s, the induced current is 4 mA, so the correct option is (A).

    Concept & Intuition

    Faraday’s law tells us that a changing magnetic flux through a coil induces an electromotive force (emf). The induced current is simply that emf divided by the coil’s resistance. Here the flux $\phi$ is given as a quadratic function of time, so its derivative is linear — easy to compute. The negative sign in Faraday’s law indicates the direction (Lenz’s law), but for the magnitude of current we only need the absolute value.

    Step-by-step solution

    1. Recall Faraday’s law The induced emf in a coil is

    $\mathcal{E} = -\frac{d\phi}{dt}$

    where $\phi$ is the magnetic flux. The negative sign tells us the direction; the magnitude is $|d\phi/dt|$.

    2. Differentiate the given flux function

    $\phi = 5t^2 + 4t + 2 \quad (\text{in } 10^{-3} \text{ weber})$

    Differentiate with respect to time:

    $\frac{d\phi}{dt} = 10t + 4 \quad (\text{in } 10^{-3} \text{ weber/s})$

    3. Evaluate at $t = 6$ seconds

    $\left.\frac{d\phi}{dt}\right|_{t=6} = 10(6) + 4 = 64 \quad (\text{in } 10^{-3} \text{ weber/s})$

    So the magnitude of the induced emf is

    $|\mathcal{E}| = 64 \times 10^{-3} \text{ V} = 0.064 \text{ V}$

    4. Apply Ohm’s law to find the induced current Resistance $R = 16\ \Omega$.

    $I = \frac{|\mathcal{E}|}{R} = \frac{0.064}{16} = 0.004 \text{ A} = 4 \text{ mA}$

    Tip

    Notice that the constant term “+2” in $\phi$ disappears upon differentiation — steady flux doesn’t induce any emf. Only the time-varying parts matter.

    Watch out

    A common mistake is forgetting the $10^{-3}$ factor from the flux unit. Always keep track of the “milli” prefix: the derivative gives $64 \times 10^{-3}$ V, not 64 V.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

11. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.The current amplification factor of a transistor in common emitter configuration is 80. If the emitter current is 2.43 mA, then the base current is (A) $15~\mu\text{A}$ (B) $1.5~\mu\text{A}$ (C) $3~\mu\text{A}$ (D) $30~\mu\text{A}$

    ›Reveal solutionSolution

    The key idea is that the current amplification factor (β) in common‑emitter configuration relates collector current to base current, and the emitter current is the sum of collector and base currents. Using β = 80 and I_E = 2.43 mA, the base current comes out to 30 μA, which corresponds to option (D).

    Concept and Intuition

    In a bipolar junction transistor (BJT) in common‑emitter configuration, the current amplification factor β (also called h_FE) is defined as the ratio of collector current to base current:

    $\beta = \frac{I_C}{I_B}$

    The three currents — emitter (I_E), base (I_B), and collector (I_C) — are related by Kirchhoff’s current law:

    $I_E = I_B + I_C$

    So if we know β and I_E, we can solve for I_B. The trick is to express I_C in terms of I_B using β, then substitute into the sum.

    Step‑by‑step reasoning

    1. Write the definition of β

    $\beta = \frac{I_C}{I_B} \quad \Rightarrow \quad I_C = \beta I_B$

    2. Use the current relation

    $I_E = I_B + I_C = I_B + \beta I_B = I_B (1 + \beta)$

    3. Solve for I_B

    $I_B = \frac{I_E}{1 + \beta}$

    4. Plug in the given numbers

    $I_E = 2.43\ \text{mA} = 2.43 \times 10^{-3}\ \text{A}, \quad \beta = 80$

    $I_B = \frac{2.43 \times 10^{-3}}{1 + 80} = \frac{2.43 \times 10^{-3}}{81}$

    5. Calculate

    $\frac{2.43}{81} = 0.03 \quad \Rightarrow \quad I_B = 0.03 \times 10^{-3}\ \text{A} = 30 \times 10^{-6}\ \text{A} = 30\ \mu\text{A}$

    Tip

    A common shortcut: since β is large (≫ 1), I_C ≈ I_E, so I_B ≈ I_E / β. Here 2.43 mA / 80 ≈ 30.4 μA — very close to the exact 30 μA. But always use the exact formula when precision matters.

    Watch out

    A classic mistake is to confuse β with α (the common‑base gain). α = I_C / I_E, and α = β/(β+1). Here β = 80 gives α ≈ 0.9877, but the problem explicitly gives β, so use the common‑emitter relation.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

12. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.A coil of resistance $16\,\Omega$ is placed with its plane perpendicular to a uniform magnetic field whose flux ($\phi$ in $10^{-3}$ weber) changes with time (t in second) as $\phi = 5t^2 + 4t + 2$. The induced current at time $t = 6$ seconds is (A) 4 mA (B) 2.12 mA (C) 34 mA (D) 74 mA

    ›Reveal solutionSolution

    The induced current is found from Faraday’s law (rate of change of flux gives emf) and Ohm’s law. At $t = 6$ s, the induced current is 4 mA.

    The central idea here is that a changing magnetic flux through a coil induces an electromotive force (emf), and that emf drives a current through the coil’s resistance. The flux is given as a function of time, so the induced emf is simply the negative time derivative of the flux. Once we have the emf, Ohm’s law gives the current.

    A common mistake is to forget that the flux is given in units of $10^{-3}$ weber — the coefficient in the expression already accounts for that, so we must treat the numbers as they are. Also, the negative sign in Faraday’s law tells us the direction of the induced current (Lenz’s law), but for the magnitude of the current we only need the absolute value of the emf.

    Let’s work through it step by step.

    1. Write Faraday’s law for the induced emf. The magnitude of the induced emf is

    $|\mathcal{E}| = \left|\frac{d\phi}{dt}\right|$

    where $\phi$ is the magnetic flux through the coil. The negative sign (Lenz’s law) determines direction; we’ll take the absolute value for the current magnitude.

    2. Differentiate the given flux function. We have $\phi = 5t^2 + 4t + 2$ (in $10^{-3}$ weber). Differentiate with respect to $t$:

    $\frac{d\phi}{dt} = 10t + 4$

    This is the rate of change of flux at any time $t$, in units of $10^{-3}$ weber per second.

    3. Evaluate at $t = 6$ seconds. Substitute $t = 6$:

    $\left.\frac{d\phi}{dt}\right|_{t=6} = 10(6) + 4 = 60 + 4 = 64$

    So the rate of change of flux is $64 \times 10^{-3}$ weber per second, i.e., $0.064$ Wb/s.

    4. Find the induced emf. The magnitude of the induced emf is exactly this value:

    $|\mathcal{E}| = 0.064 \text{ V}$

    (since 1 Wb/s = 1 V).

    5. Apply Ohm’s law to get the induced current. The coil has resistance $R = 16\,\Omega$. The current is

    $I = \frac{|\mathcal{E}|}{R} = \frac{0.064}{16} = 0.004 \text{ A}$

    Convert to milliamperes: $0.004 \text{ A} = 4 \text{ mA}$.

    Watch out

    A common slip is to forget that the flux is already in $10^{-3}$ weber, so the derivative $10t+4$ is also in those units. If you mistakenly treat the numbers as plain webers, you’d get $64$ V instead of $0.064$ V — leading to a wildly wrong current of $4$ A. Always check the units given in the problem.

    Tip

    Notice that the constant term $+2$ in $\phi$ disappears on differentiation — a constant flux contributes nothing to the induced emf. Only the time-varying part matters.

    ✓Final answer

    The induced current at $t = 6$ s is 4 mA, which corresponds to option (A).