Q. What is the impedance of a series RLC circuit at resonance?
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Solution
At resonance, the impedance of a series RLC circuit is equal to the resistance R.
Correct Answer: A — R
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Q. What is the induced EMF in a loop of wire if the magnetic field through the loop changes at a rate of dB/dt?
A.
-A(dB/dt)
B.
A(dB/dt)
C.
-L(dB/dt)
D.
L(dB/dt)
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Solution
The induced EMF (ε) is given by Faraday's law as ε = -A(dB/dt), where A is the area of the loop.
Correct Answer: A — -A(dB/dt)
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Q. What is the induced EMF in a loop of wire if the magnetic flux through it changes at a rate of dΦ/dt?
A.
-dΦ/dt
B.
dΦ/dt
C.
Φ
D.
0
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Solution
According to Faraday's law of electromagnetic induction, the induced EMF is given by ε = -dΦ/dt.
Correct Answer: A — -dΦ/dt
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Q. What is the induced EMF in a loop of wire when the magnetic field changes at a rate of 5 T/s?
A.
0 V
B.
5 V
C.
10 V
D.
15 V
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Solution
The induced EMF (ε) is given by Faraday's law of electromagnetic induction: ε = -dΦ/dt. If the change in magnetic field is 5 T/s, then ε = 5 V.
Correct Answer: B — 5 V
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Q. What is the induced EMF in a loop of wire when the magnetic field through it changes at a rate of dB/dt?
A.
-dB/dt
B.
dB/dt
C.
μ₀dB/dt
D.
0
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Solution
According to Faraday's law of electromagnetic induction, the induced EMF is given by ε = -dΦ/dt, where Φ is the magnetic flux.
Correct Answer: A — -dB/dt
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Q. What is the induced EMF in a loop of wire when the magnetic field through it changes at a rate of 5 T/s?
A.
0 V
B.
5 V
C.
10 V
D.
15 V
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Solution
The induced EMF (ε) is given by Faraday's law of electromagnetic induction: ε = -dΦ/dt. If the rate of change of magnetic field is 5 T/s, then ε = 5 V.
Correct Answer: B — 5 V
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Q. What is the integral form of Ampere's Law?
A.
∮B·dl = μ₀I_enclosed
B.
∮E·dl = -dΦ/dt
C.
∮F·dl = m*a
D.
∮V·dl = Q/C
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Solution
The integral form of Ampere's Law states that the line integral of the magnetic field B around a closed loop is equal to μ₀ times the enclosed current I.
Correct Answer: A — ∮B·dl = μ₀I_enclosed
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Q. What is the latent heat of fusion for ice?
A.
334 J/g
B.
2260 J/g
C.
4190 J/g
D.
1000 J/g
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Solution
The latent heat of fusion for ice is approximately 334 J/g.
Correct Answer: A — 334 J/g
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Q. What is the latent heat of fusion?
A.
Heat required to change a solid to a liquid
B.
Heat required to change a liquid to a gas
C.
Heat required to change a gas to a solid
D.
Heat required to change a liquid to a solid
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Solution
Latent heat of fusion is the heat required to change a solid into a liquid at its melting point.
Correct Answer: A — Heat required to change a solid to a liquid
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Q. What is the magnetic field at a distance r from an infinitely long straight wire carrying current I?
A.
μ₀I/2πr
B.
μ₀I/4πr
C.
μ₀I/πr
D.
0
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Solution
Using Ampere's Law, B = μ₀I/2πr for an infinitely long straight wire.
Correct Answer: A — μ₀I/2πr
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Q. What is the magnetic field at a point due to a long straight current-carrying conductor using Biot-Savart Law?
A.
μ₀I/(2πr)
B.
μ₀I/(4πr²)
C.
μ₀I/(2r)
D.
μ₀I/(4r)
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Solution
The magnetic field B at a distance r from a long straight conductor carrying current I is given by B = μ₀I/(2πr) according to Biot-Savart Law.
Correct Answer: A — μ₀I/(2πr)
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Q. What is the magnetic field at a point on the axis of a circular loop of radius R carrying current I?
A.
μ₀I/(2R)
B.
μ₀I/(4R)
C.
μ₀I/(2R²)
D.
μ₀I/(2√2R)
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Solution
The magnetic field on the axis of a circular loop is given by B = (μ₀I/(2R)) * (1/(1 + (z/R)²)^(3/2)) where z is the distance along the axis.
Correct Answer: D — μ₀I/(2√2R)
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Q. What is the magnetic field at a point on the axis of a circular loop of radius R carrying a current I at a distance x from the center?
A.
(μ₀I)/(2R) * (R²/(R²+x²)^(3/2))
B.
(μ₀I)/(4R) * (R²/(R²+x²)^(3/2))
C.
(μ₀I)/(2R) * (1/(R²+x²)^(3/2))
D.
(μ₀I)/(4R) * (1/(R²+x²)^(3/2))
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Solution
The magnetic field on the axis of a circular loop is given by B = (μ₀I)/(2R) * (R²/(R²+x²)^(3/2)).
Correct Answer: A — (μ₀I)/(2R) * (R²/(R²+x²)^(3/2))
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Q. What is the magnetic field at the center of a circular loop of radius R carrying current I?
A.
μ₀I/2R
B.
μ₀I/R
C.
μ₀I/4R
D.
μ₀I/πR
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Solution
The magnetic field at the center of a circular loop is given by B = μ₀I/2R.
Correct Answer: A — μ₀I/2R
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Q. What is the magnetic field at the center of a circular loop of radius R carrying a current I?
A.
μ₀I/(2R)
B.
μ₀I/(4R)
C.
μ₀I/(R)
D.
μ₀I/(8R)
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Solution
The magnetic field at the center of a circular loop is given by B = μ₀I/(2R).
Correct Answer: A — μ₀I/(2R)
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Q. What is the magnetic field at the center of a square loop of side a carrying current I?
A.
μ₀I/4a
B.
μ₀I/2a
C.
μ₀I/a
D.
μ₀I/8a
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Solution
The magnetic field at the center of a square loop is B = μ₀I/4a.
Correct Answer: A — μ₀I/4a
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Q. What is the magnetic field at the midpoint of a wire carrying current I in opposite directions?
A.
Zero
B.
μ₀I/2
C.
μ₀I
D.
Depends on distance
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Solution
At the midpoint, the magnetic fields due to the two currents cancel each other out, resulting in zero net magnetic field.
Correct Answer: A — Zero
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Q. What is the magnetic field due to a circular loop of radius R carrying current I at a point on its axis at a distance x from the center?
A.
μ₀I/(2R)
B.
μ₀I/(2(x² + R²)^(3/2))
C.
μ₀I/(4πR)
D.
μ₀I/(x² + R²)
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Solution
The magnetic field at a point on the axis of a circular loop at a distance x from the center is given by B = μ₀I/(2(x² + R²)^(3/2)).
Correct Answer: B — μ₀I/(2(x² + R²)^(3/2))
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Q. What is the magnetic field due to a current I flowing through a straight wire at a distance d?
A.
μ₀I/(2πd)
B.
μ₀I/(4πd²)
C.
μ₀I/(d)
D.
μ₀I/(2d)
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Solution
The magnetic field B at a distance d from a straight wire carrying current I is given by B = μ₀I/(2πd).
Correct Answer: A — μ₀I/(2πd)
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Q. What is the magnetic field due to a current loop at a point on its axis?
A.
μ₀I/2R
B.
μ₀I/4R
C.
μ₀I/2R²
D.
μ₀I/4R²
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Solution
Using Ampere's Law, B = μ₀I/2R² at a point on the axis of a current loop.
Correct Answer: C — μ₀I/2R²
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Q. What is the magnetic field due to a long straight wire carrying current I at a distance r from the wire?
A.
μ₀I/2πr
B.
μ₀I/4πr
C.
μ₀I/πr
D.
μ₀I/8πr
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Solution
The magnetic field due to a long straight wire is given by B = (μ₀I)/(2πr).
Correct Answer: A — μ₀I/2πr
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Q. What is the magnetic field due to a magnetic dipole at a point along its axial line?
A.
(μ₀/4π) * (2m/r³)
B.
(μ₀/4π) * (m/r³)
C.
(μ₀/4π) * (m/r²)
D.
(μ₀/4π) * (m/r⁴)
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Solution
The magnetic field due to a magnetic dipole at a point along its axial line is given by B = (μ₀/4π) * (2m/r³).
Correct Answer: A — (μ₀/4π) * (2m/r³)
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Q. What is the magnetic field due to a solenoid of length L, carrying n turns per unit length and current I?
A.
μ₀nI
B.
μ₀nI/L
C.
μ₀nI/2
D.
μ₀nI/L²
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Solution
The magnetic field inside a long solenoid is B = μ₀nI.
Correct Answer: A — μ₀nI
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Q. What is the magnetic field due to a straight conductor at a point 1 meter away carrying a current of 5 A?
A.
0.01 T
B.
0.02 T
C.
0.03 T
D.
0.04 T
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Solution
Using Biot-Savart Law, B = μ₀I/(2πr) = (4π × 10^-7 Tm/A)(5 A)/(2π(1 m)) = 0.01 T.
Correct Answer: B — 0.02 T
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Q. What is the magnetic field due to a straight wire at a distance of 0.5 m carrying a current of 10 A?
A.
0.4 μT
B.
0.2 μT
C.
0.1 μT
D.
0.8 μT
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Solution
Using B = μ₀I/(2πr), substituting μ₀ = 4π × 10⁻⁷ Tm/A, I = 10 A, and r = 0.5 m gives B = 0.4 μT.
Correct Answer: A — 0.4 μT
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Q. What is the magnetic field inside a hollow conductor carrying current?
A.
Zero
B.
Uniform
C.
Varies with distance
D.
Depends on the current
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Solution
According to Ampere's law, the magnetic field inside a hollow conductor carrying current is zero.
Correct Answer: A — Zero
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Q. What is the magnetic field inside a hollow cylindrical shell carrying current I?
A.
0
B.
μ₀I/2πR
C.
μ₀I/4πR
D.
μ₀I/πR
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Solution
Inside a hollow cylindrical shell, the magnetic field is zero.
Correct Answer: A — 0
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Q. What is the magnetic field inside a long solenoid carrying current I?
A.
Zero
B.
μ₀nI
C.
μ₀I/n
D.
μ₀I/2
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Solution
The magnetic field inside a long solenoid is given by B = μ₀nI, where n is the number of turns per unit length.
Correct Answer: B — μ₀nI
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Q. What is the magnetic field inside a long solenoid carrying current?
A.
Zero
B.
Uniform and parallel to the axis
C.
Varies with distance
D.
Depends on the current only
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Solution
The magnetic field inside a long solenoid is uniform and parallel to the axis of the solenoid.
Correct Answer: B — Uniform and parallel to the axis
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Q. What is the magnetic field inside a long solenoid with n turns per unit length carrying a current I?
A.
μ₀nI
B.
μ₀I/n
C.
μ₀I/2n
D.
μ₀I/4n
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Solution
The magnetic field inside a long solenoid is given by B = μ₀nI.
Correct Answer: A — μ₀nI
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