Unit 14: Electromagnetic Induction and Alternating Currents — Online MCQ Test
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Practice Unit 14: Electromagnetic Induction and Alternating Currents with a free chapter-wise online MCQ test.
This chapter covers: Electromagnetic induction: Faraday's law induced emf and current Lenz’s law eddy currents self and mutual inductance. Alternating currents peak and RMS value of alternating current....
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Unit 14: Electromagnetic Induction and Alternating Currents — Important Questions & Answers
A capacitor C is first charged fully with potential difference of V0 and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In t s 25% of the initial energy in the capacitor is transferred to the inductor. The value of t is ____ s.
- A. (LC^1/2 * π)/3
- B. (LC^1/2 * π)/6
- C. (LC^1/2 * π)/2
- D. (LC^1/2 * π)/2
Answer: B. (LC^1/2 * π)/6
Energy in inductor U_L = 25% of U_initial = 1/4 * (1/2)CV0^2. U_L = (1/2)Li^2. Current in LC circuit follows i = I_max*sin(ωt). Energy transferred corresponds to sin^2(ωt) = 0.25, so sin(ωt) = 0.5. Thus ωt = π/6. With ω = 1/sqrt(LC), t = sqrt(LC) * π/6.
Energy in inductor U_L = 25% of U_initial = 1/4 * (1/2)CV0^2. U_L = (1/2)Li^2. Current in LC circuit follows i = I_max*sin(ωt). Energy transferred corresponds to sin^2(ωt) = 0.25, so sin(ωt) = 0.5. Thus ωt = π/6. With ω = 1/sqrt(LC), t = sqrt(LC) * π/6.
Figure shows the circuit that contains three resistances (9Ω each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is ____ A.
- A. 1
- B. zero
- C. 3
- D. 2
Answer: A. 1
At the moment the switch is closed (t=0), the inductors act as an open circuit (infinite resistance). The circuit simplifies to a single resistor branch carrying the current. Given V=9V (implied by circuit structure) and R=9Ω, I = V/R = 9/9 = 1A.
At the moment the switch is closed (t=0), the inductors act as an open circuit (infinite resistance). The circuit simplifies to a single resistor branch carrying the current. Given V=9V (implied by circuit structure) and R=9Ω, I = V/R = 9/9 = 1A.
A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by 30° from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is α mm. Find α.
- A. 7
- B. 14
- C. 21
- D. 28
Answer: B. 14
The induced EMF is E = BAω sin(ωt). Given E = 15.4 * 10^-3 V, B = 0.5 T, ω = 100 rad/s, and ωt = 30°. Thus 15.4 * 10^-3 = 0.5 * (π * r^2) * 100 * sin(30°). Solving for r, we find r = 14 mm.
The induced EMF is E = BAω sin(ωt). Given E = 15.4 * 10^-3 V, B = 0.5 T, ω = 100 rad/s, and ωt = 30°. Thus 15.4 * 10^-3 = 0.5 * (π * r^2) * 100 * sin(30°). Solving for r, we find r = 14 mm.
XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is l. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > l) and mass m is made to slide down from rest under the gravity. The terminal speed acquired by the rod is ____ m/s.
- A. 2mgR / (B^2 l^2)
- B. 8mgR / (B^2 l^2)
- C. 2mgR / (B^2 L^2)
- D. mgR / (B^2 l^2)
Answer: D. mgR / (B^2 l^2)
Terminal velocity is reached when gravitational force equals magnetic force (F_mag = ilB = (Bv l / R) * lB). Set mg = (B^2 l^2 v) / R. Solving for v gives v = mgR / (B^2 l^2).
Terminal velocity is reached when gravitational force equals magnetic force (F_mag = ilB = (Bv l / R) * lB). Set mg = (B^2 l^2 v) / R. Solving for v gives v = mgR / (B^2 l^2).
मुक्त आकाश में एक विद्युतचुंबकीय तरंग के विद्युत क्षेत्र $E$ तथा चुंबकीय क्षेत्र $B$ से सम्बन्धित ऊर्जा घनत्व ($\varepsilon_0$-मुक्त आकाश की विद्युतशीलता, $\mu_0$ मुक्त आकाश की चुंबकीयशीलता):
- A. $U_E = \frac{\varepsilon_0 E^2}{2},\ U_B = \frac{\mu_0 B^2}{2}$
- B. $U_E = \frac{E^2}{2\varepsilon_0},\ U_B = \frac{B^2}{2\mu_0}$
- C. $U_E = \frac{\varepsilon_0 E^2}{2},\ U_B = \frac{B^2}{2\mu_0}$
- D. $U_E = \frac{E^2}{2\varepsilon_0},\ U_B = \frac{\mu_0 B^2}{2}$
Answer: C. $U_E = \frac{\varepsilon_0 E^2}{2},\ U_B = \frac{B^2}{2\mu_0}$
In free space, the electric energy density is $$U_E = \frac{1}{2}\varepsilon_0 E^2,$$ and the magnetic energy density is $$U_B = \frac{B^2}{2\mu_0}.$$ So option C is correct.
In free space, the electric energy density is $$U_E = \frac{1}{2}\varepsilon_0 E^2,$$ and the magnetic energy density is $$U_B = \frac{B^2}{2\mu_0}.$$ So option C is correct.