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UPMSP Intermediate Physics - Code 151 study materials
UPMSP Intermediate Physics - Code 151 study resources

UPMSP Intermediate Physics - Code 151 practice questions

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Question 1Electrostatics

Two point charges of +4 μC and −4 μC are separated by 0.2 m in air. What is the magnitude of the force between them?

  1. 1.8 N
  2. 0.36 N
  3. 3.6 N Correct answer.
  4. 7.2 N

Explanation

Coulomb's law gives F = kq₁q₂/r² with k = 9×10⁹ N m² C⁻². Substituting q₁ = q₂ = 4×10⁻⁶ C and r = 0.2 m gives F = (9×10⁹ × 16×10⁻¹²)/0.04 = 3.6 N. The opposite signs mean the force is attractive, but the magnitude is what is asked.

Question 2Electrostatics

Which physical quantity is defined as the electric flux passing through a surface per unit area taken normal to the surface?

  1. Electric potential
  2. Electric dipole moment
  3. Surface charge density
  4. Electric field intensity Correct answer.

Explanation

Electric field intensity is the flux crossing unit area held perpendicular to the field lines, so E = φ/A for a normal surface. This definition makes the field a vector quantity measured in N C⁻¹ or V m⁻¹. Potential, charge density and dipole moment all have different definitions and units.

Question 3Electrostatics

A charge of 5 μC is placed at a point where the electric potential is 200 V. What is the potential energy of the charge at that point?

  1. 1×10⁻³ J Correct answer.
  2. 4×10⁻⁴ J
  3. 1×10⁻⁶ J
  4. 2.5×10⁻⁸ J

Explanation

The potential energy of a charge q at a point of potential V is U = qV. Here U = 5×10⁻⁶ C × 200 V = 1×10⁻³ J. Potential energy is a scalar, so only the magnitude of the charge enters the product.

Question 4Electrostatics

The electric field inside the cavity of a charged hollow conductor, in electrostatic equilibrium, is

  1. directed radially outward from the cavity centre
  2. equal to the field at its outer surface
  3. maximum at the centre of the cavity
  4. zero at every point inside the cavity Correct answer.

Explanation

In electrostatic equilibrium the free charges on a conductor reside on its outer surface, and the field within the material and any enclosed empty cavity is zero. This is the basis of electrostatic shielding. The absence of charge on the inner surface of an empty cavity is a consequence of Gauss's law.

Question 5Electrostatics

A parallel plate capacitor of capacitance C is charged to potential V and then disconnected from the battery. If the plate separation is doubled while the charge stays constant, the stored energy becomes

  1. half the original
  2. four times the original
  3. unchanged
  4. twice the original Correct answer.

Explanation

With charge fixed, energy is U = Q²/(2C), and doubling the separation halves the capacitance since C = ε₀A/d. Halving C therefore doubles the stored energy. The extra energy comes from the mechanical work done in pulling the plates apart against their attraction.

Question 6Electrostatics

The electric potential at a distance r from an isolated point charge Q is proportional to

  1. Q/r²
  2. Q/r Correct answer.
  3. Q²/r
  4. Q r

Explanation

The potential of a point charge is V = Q/(4πε₀r), so it varies as Q/r and falls off more slowly than the field. The field, by contrast, varies as Q/r². Potential is a scalar, so its sign follows the sign of Q.

Question 7Electrostatics

Two capacitors of 3 μF and 6 μF are joined in series across a 12 V supply. What is the charge on each capacitor?

  1. 24 μC Correct answer.
  2. 12 μC
  3. 72 μC
  4. 36 μC

Explanation

In series the equivalent capacitance is (3×6)/(3+6) = 2 μF, and the same charge flows through both. The charge is Q = CV = 2 μF × 12 V = 24 μC. Each capacitor carries this same charge even though the voltages across them differ.

Question 8Electrostatics

An electric dipole of moment p is placed in a uniform electric field E with its axis making an angle θ with the field. The torque on the dipole is

  1. pE sin θ Correct answer.
  2. pE cos θ
  3. pE tan θ
  4. pE

Explanation

The torque on a dipole is the vector product of p and E, whose magnitude is pE sin θ. It is maximum when the dipole lies perpendicular to the field and zero when aligned with it. The potential energy, in contrast, involves cos θ.

Question 9Electrostatics

A charge Q is placed at the centre of a cube of side a. What is the electric flux through one face of the cube?

  1. Q/ε₀
  2. Q/(4πε₀a²)
  3. Q/(8ε₀)
  4. Q/(6ε₀) Correct answer.

Explanation

By Gauss's law the total flux through the closed cube is Q/ε₀. Symmetry about the centre distributes this equally over the six identical faces, so each face carries Q/(6ε₀). The side length does not appear because the total flux depends only on the enclosed charge.

Question 10Electrostatics

The electric field just outside the surface of a charged conductor in air is 3×10⁵ N C⁻¹. What is the surface charge density on the conductor?

  1. 8.85×10⁻¹² C m⁻²
  2. 5.31×10⁻⁶ C m⁻²
  3. 3.32×10⁻¹⁷ C m⁻²
  4. 2.66×10⁻⁶ C m⁻² Correct answer.

Explanation

Just outside a conductor E = σ/ε₀, so σ = ε₀E = 8.85×10⁻¹² × 3×10⁵ ≈ 2.66×10⁻⁶ C m⁻². The field is normal to the surface there. This relation follows from applying Gauss's law to a small pillbox straddling the surface.

Question 11Electrostatics

A 2 μF capacitor is charged to 100 V and then connected in parallel with an uncharged 3 μF capacitor. What is the common potential?

  1. 100 V
  2. 40 V Correct answer.
  3. 50 V
  4. 60 V

Explanation

The initial charge is 2 μF × 100 V = 200 μC and is shared between the two capacitors when they are joined in parallel. The total capacitance becomes 5 μF, so the common potential is 200 μC / 5 μF = 40 V. Charge is conserved during sharing.

Question 12Electrostatics

The work done in moving a charge of 2 μC between two points whose potentials differ by 50 V is

  1. 0.1 mJ
  2. 100 μJ Correct answer.
  3. 25 μJ
  4. 2.5 mJ

Explanation

The work done by the external agent equals q times the potential difference, W = qΔV. Here W = 2×10⁻⁶ C × 50 V = 1×10⁻⁴ J, that is 100 μJ. The path taken between the two points does not affect this value because the electrostatic field is conservative.

Question 13Electrostatics

A hollow metallic sphere of radius R carries a charge Q. How does the electric field vary with distance r from the centre for r > R?

  1. It varies as 1/r inside and as 1/r² outside
  2. It is the same as if all the charge were concentrated at the centre Correct answer.
  3. It falls to zero at the outer surface and stays zero beyond
  4. It is zero everywhere outside the sphere

Explanation

For points outside a uniformly charged spherical shell, Gauss's law with a sphere of radius r > R encloses the whole charge Q, giving E = Q/(4πε₀r²). This is identical to the field of a point charge Q at the centre. Inside the shell (r < R) the field is zero.

Question 14Electrostatics

Two identical small spheres each of mass m and charge q hang from a common point by insulating threads of length L. At equilibrium the threads make a small angle with the vertical. The tension in each thread is

  1. mg cos θ
  2. mg tan θ
  3. qE sin θ
  4. mg / cos θ Correct answer.

Explanation

Resolving vertically, the upward component of tension T cos θ balances the weight mg, so T = mg/cos θ. The horizontal component T sin θ balances the electric repulsion. This resolution is the standard way to find tension in such equilibrium problems.

Question 15Electrostatics

A dielectric slab of dielectric constant K completely fills the space between the plates of a charged capacitor that remains connected to a battery. The energy stored becomes

  1. unchanged
  2. 1/K times the original
  3. K² times the original
  4. K times the original Correct answer.

Explanation

With the battery connected, the potential V stays fixed while the capacitance rises to KC. The energy U = ½CV² therefore becomes K times its previous value. The battery supplies the extra energy, and the dielectric is drawn into the gap.

Question 16Electrostatics

A parallel plate capacitor has plates of area A separated by d. A slab of thickness d/2 and dielectric constant K is inserted between the plates, filling half the gap. The new capacitance is

  1. (K+1)ε₀A / (2d)
  2. Kε₀A / d
  3. ε₀A / (d(K+1))
  4. 2Kε₀A / ((K+1)d) Correct answer.

Explanation

The air layer and the dielectric layer behave as two capacitors in series, each of area A and thickness d/2. Their equivalent capacitance is found from 1/C = d/(2ε₀A) + d/(2Kε₀A), giving C = 2Kε₀A/((K+1)d). This lies between the air-filled value and the fully filled value.

Question 17Current Electricity

किसी चालक के प्रतिरोध R का सूत्र, जहाँ ρ विशिष्ट प्रतिरोध, l लंबाई और A अनुप्रस्थ काट का क्षेत्रफल है, क्या है?

  1. R = ρl/A Correct answer.
  2. R = ρA/l
  3. R = lA/ρ
  4. R = ρ/(lA)

Explanation

प्रतिरोध लंबाई के समानुपाती और अनुप्रस्थ काट के क्षेत्रफल के व्युत्क्रमानुपाती होता है, इसलिए R = ρl/A। यहाँ ρ पदार्थ की विशिष्ट प्रतिरोधकता है जो ताप पर निर्भर करती है। लंबाई दोगुनी करने पर प्रतिरोध दोगुना और क्षेत्रफल दोगुना करने पर आधा हो जाता है।

Question 18Current Electricity

श्रेणीक्रम में जुड़े तीन प्रतिरोधों 2 Ω, 3 Ω और 5 Ω का तुल्य प्रतिरोध कितना होगा?

  1. 30 Ω
  2. 10 Ω Correct answer.
  3. 5 Ω
  4. 0.97 Ω

Explanation

श्रेणीक्रम संयोजन में तुल्य प्रतिरोध सभी प्रतिरोधों के साधारण योग के बराबर होता है, अर्थात R = 2 + 3 + 5 = 10 Ω। श्रेणीक्रम में हर प्रतिरोध से समान धारा गुजरती है और वोल्टता विभाजित होती है।

Question 19Current Electricity

किसी सेल के विद्युत वाहक बल (emf) की परिभाषा के अनुसार यह किसके बराबर होता है?

  1. एकांक आवेश को पूरे परिपथ में घुमाने में सेल द्वारा किया गया कार्य Correct answer.
  2. सेल के भीतर आंतरिक प्रतिरोध पर होने वाली वोल्टता पतन
  3. सेल के सिरों के बीच की वोल्टता जब परिपथ में धारा बह रही हो
  4. सेल से प्राप्त अधिकतम धारा और आंतरिक प्रतिरोध का गुणनफल

Explanation

विद्युत वाहक बल वह कार्य है जो सेल एकांक धनावेश को एक इलेक्ट्रोड से दूसरे तक पहुँचाने अर्थात पूरे परिपथ में घुमाने में करता है। इसका मात्रक वोल्ट है और यह सेल की रासायनिक प्रकृति पर निर्भर करता है। यह टर्मिनल वोल्टता से भिन्न होता है क्योंकि धारा बहने पर आंतरिक प्रतिरोध के कारण कुछ वोल्टता गिर जाती है।

Question 20Current Electricity

किसी धात्विक चालक का ताप बढ़ाने पर उसका प्रतिरोध सामान्यतः क्या होता है?

  1. बढ़ जाता है Correct answer.
  2. पहले घटता फिर स्थिर रहता है
  3. अपरिवर्तित रहता है
  4. घट जाता है

Explanation

धातुओं में ताप बढ़ने पर जालक के आयनों का कंपन बढ़ जाता है, जिससे इलेक्ट्रॉनों का टकराव अधिक होता है और प्रतिरोध बढ़ जाता है। इसीलिए धातुओं का ताप गुणांक धनात्मक होता है।

Question 21Current Electricity

एक तार को खींचकर उसकी लंबाई दोगुनी कर दी जाती है, आयतन समान रहता है। नया प्रतिरोध पुराने प्रतिरोध का कितना गुना होगा?

  1. 2 गुना
  2. आधा
  3. 4 गुना Correct answer.
  4. 8 गुना

Explanation

आयतन नियत रहने पर लंबाई दोगुनी करने से अनुप्रस्थ काट का क्षेत्रफल आधा हो जाता है। चूँकि R = ρl/A, लंबाई दोगुनी और क्षेत्रफल आधा होने से प्रतिरोध 2 × 2 = 4 गुना हो जाता है।

Question 22Current Electricity

4 Ω और 6 Ω के दो प्रतिरोध समांतरक्रम में जोड़े गए हैं। इस संयोजन का तुल्य प्रतिरोध क्या है?

  1. 10 Ω
  2. 2.4 Ω Correct answer.
  3. 5 Ω
  4. 1.5 Ω

Explanation

समांतरक्रम में 1/R = 1/4 + 1/6 = 3/12 + 2/12 = 5/12, इसलिए R = 12/5 = 2.4 Ω। समांतर संयोजन का तुल्य प्रतिरोध सबसे छोटे प्रतिरोध से भी कम होता है।

Question 23Current Electricity

किसी सेल का emf 12 V और आंतरिक प्रतिरोध 2 Ω है। इसे 4 Ω के बाह्य प्रतिरोध से जोड़ने पर परिपथ में धारा कितनी होगी?

  1. 1 A
  2. 6 A
  3. 2 A Correct answer.
  4. 3 A

Explanation

पूरे परिपथ का कुल प्रतिरोध बाह्य और आंतरिक प्रतिरोधों का योग होता है, अर्थात 4 + 2 = 6 Ω। इसलिए धारा I = ε/(R + r) = 12/6 = 2 A।

Question 24Current Electricity

किरचॉफ का संधि नियम (junction rule) किस संरक्षण सिद्धांत का सीधा परिणाम है?

  1. संवेग का संरक्षण
  2. चुंबकीय फ्लक्स का संरक्षण
  3. आवेश का संरक्षण Correct answer.
  4. ऊर्जा का संरक्षण

Explanation

संधि नियम कहता है कि किसी संधि पर मिलने वाली धाराओं का बीजगणितीय योग शून्य होता है, अर्थात संधि में आने वाला आवेश बाहर जाने वाले आवेश के बराबर होता है। यह आवेश संरक्षण का ही कथन है। ऊर्जा संरक्षण से लूप नियम जुड़ा है।

Question 25Current Electricity

किसी सेल के लिए V = ε − Ir संबंध में, यदि धारा I बढ़ाई जाए तो टर्मिनल वोल्टता V पर क्या प्रभाव पड़ता है?

  1. V अपरिवर्तित रहती है
  2. V घटती है Correct answer.
  3. V बढ़ती है
  4. V पहले बढ़ती फिर घटती है

Explanation

संबंध V = ε − Ir बताता है कि आंतरिक प्रतिरोध पर वोल्टता पतन Ir धारा के समानुपाती है। अतः धारा बढ़ने पर Ir बढ़ता है और टर्मिनल वोल्टता घट जाती है। emf ε सेल की स्थिर रासायनिक विशेषता है।

Question 26Current Electricity

किसी धात्विक चालक में धारा घनत्व J और अपवाह वेग vd के बीच सही संबंध कौन सा है, जहाँ n इलेक्ट्रॉन घनत्व और e आवेश है?

  1. J = ne/vd
  2. J = nvd/e
  3. J = nevd Correct answer.
  4. J = vd/(ne)

Explanation

धारा I = neAvd होती है, जहाँ A अनुप्रस्थ काट का क्षेत्रफल है। धारा घनत्व J = I/A रखने पर J = nevd प्राप्त होता है। इससे स्पष्ट है कि धारा घनत्व अपवाह वेग के समानुपाती है।

Question 27Current Electricity

एक परिपथ में 6 V की बैटरी (आंतरिक प्रतिरोध नगण्य) 2 Ω, 3 Ω और 6 Ω के तीन प्रतिरोधों के समांतर संयोजन से जुड़ी है। बैटरी से निकलने वाली कुल धारा कितनी होगी?

  1. 11 A
  2. 1 A
  3. 3 A
  4. 6 A Correct answer.

Explanation

समांतर संयोजन का तुल्य प्रतिरोध 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1, अतः R = 1 Ω। ओम के नियम से कुल धारा I = V/R = 6/1 = 6 A। प्रत्येक शाखा में धारा अलग-अलग होगी पर योग 6 A ही रहेगा।

Question 28Current Electricity

किसी सेल के लिए खींचा गया V-I ग्राफ एक ऋणात्मक ढाल वाली सरल रेखा है। इस रेखा का ढाल और अक्ष पर अंतःखंड क्रमशः क्या दर्शाते हैं?

  1. ढाल बाह्य प्रतिरोध और अंतःखंड धारा दर्शाता है
  2. ढाल आंतरिक प्रतिरोध और अंतःखंड emf दर्शाता है Correct answer.
  3. ढाल emf और अंतःखंड आंतरिक प्रतिरोध दर्शाता है
  4. ढाल शक्ति और अंतःखंड आवेश दर्शाता है

Explanation

V = ε − Ir की तुलना y = c + mx से करने पर ढाल −r अर्थात आंतरिक प्रतिरोध का ऋणात्मक मान मिलता है और V-अक्ष पर अंतःखंड ε होता है। इस प्रकार ग्राफ से सेल का emf और आंतरिक प्रतिरोध दोनों ज्ञात किए जा सकते हैं।

Question 29Current Electricity

एक मीटर सेतु (metre bridge) में संतुलन बिंदु तार के 40 cm पर प्राप्त होता है, जब दाहिनी गैप में 6 Ω का प्रतिरोध है। बाईं गैप में रखा अज्ञात प्रतिरोध कितना है?

  1. 2.4 Ω
  2. 4 Ω Correct answer.
  3. 15 Ω
  4. 9 Ω

Explanation

मीटर सेतु में संतुलन पर R/S = l/(100 − l), जहाँ l बाईं ओर की संतुलन लंबाई है। यहाँ R/6 = 40/60 = 2/3, अतः R = 6 × 2/3 = 4 Ω। संतुलन की स्थिति में धारामापी से कोई धारा नहीं बहती।

Question 30Current Electricity

किसी सेल से अधिकतम शक्ति बाह्य प्रतिरोध R में स्थानांतरित होती है जब R आंतरिक प्रतिरोध r के बराबर होता है। इस स्थिति में सेल से बाह्य परिपथ को दी गई शक्ति का व्यंजक क्या है?

  1. ε²/r
  2. ε²/(4r) Correct answer.
  3. ε²/(2r)
  4. 2ε²/r

Explanation

अधिकतम शक्ति स्थानांतरण की शर्त R = r पर कुल प्रतिरोध 2r होता है और धारा I = ε/(2r)। बाह्य प्रतिरोध में शक्ति P = I²R = (ε/2r)² × r = ε²/(4r) होती है। इसी कारण उपयोगी शक्ति अधिकतम होने पर दक्षता केवल 50 प्रतिशत रहती है।

Question 31Magnetic Effects of Current and Magnetism

A long straight conductor carries a steady current I. At a perpendicular distance r from the wire, the magnetic field magnitude B is given by which expression? (μ₀ is the permeability of free space.)

  1. B = μ₀Ir/(2π)
  2. B = μ₀I²/(2πr)
  3. B = μ₀I/(4πr²)
  4. B = μ₀I/(2πr) Correct answer.

Explanation

Ampère's circuital law applied to a circular path of radius r around the wire gives B(2πr) = μ₀I, so B = μ₀I/(2πr). The field falls off as 1/r, not as 1/r² which is the Coulomb-type dependence for a point charge. The result is independent of the medium only through μ₀; inside a magnetic material μ₀ is replaced by μ.

Question 32Magnetic Effects of Current and Magnetism

According to the Biot–Savart law, the magnitude of the magnetic field dB produced by a current element I dl at a point whose position vector from the element is r is proportional to which quantity?

  1. I dl cosθ / r
  2. I dl / r³
  3. I dl sinθ / r² Correct answer.
  4. I dl tanθ / r²

Explanation

The Biot–Savart law states dB = (μ₀/4π)(I dl sinθ)/r², where θ is the angle between the current element and the position vector. The sine dependence means the field is zero when the element points directly toward or away from the point. The inverse-square dependence on r matches an element acting as a point source.

Question 33Magnetic Effects of Current and Magnetism

A charged particle of charge q moves with velocity v in a region where the magnetic field is B. The magnetic force on the particle is given by which relation?

  1. F = q(v · B)
  2. F = q(v × B) Correct answer.
  3. F = q(v + B)
  4. F = qB/v

Explanation

The magnetic Lorentz force is F = q(v × B), a vector cross product whose direction is perpendicular to both v and B. Because the force is always perpendicular to the velocity, the magnetic field does no work on the particle and cannot change its speed. The magnitude is qvB sinθ, where θ is the angle between v and B.

Question 34Magnetic Effects of Current and Magnetism

The magnetic field at the centre of a circular coil of radius R carrying current I and having N turns is given by which expression?

  1. B = μ₀NI/(4R)
  2. B = μ₀I/(2πR)
  3. B = μ₀NI R
  4. B = μ₀NI/(2R) Correct answer.

Explanation

For a single circular loop, B at the centre is μ₀I/(2R); with N turns the contributions add, giving B = μ₀NI/(2R). The field is directed along the axis of the coil as given by the right-hand rule. Increasing the number of turns or decreasing the radius both strengthen the field at the centre.

Question 35Magnetic Effects of Current and Magnetism

A current-carrying conductor is placed in a uniform magnetic field. The magnetic force on the conductor is maximum when the angle between the conductor and the field is:

  1. 180°
  2. 0°
  3. 45°
  4. 90° Correct answer.

Explanation

The force on a straight conductor of length L is F = BIL sinθ, which is largest when sinθ = 1, that is, when θ = 90°. When the conductor lies along the field (θ = 0° or 180°) the force is zero. At 45° the force is only about 0.707 of its maximum value.

Question 36Magnetic Effects of Current and Magnetism

Two long parallel straight wires carry currents in the same direction. The nature of the force between them is:

  1. Attractive Correct answer.
  2. Alternating between attractive and repulsive
  3. Zero because the fields cancel everywhere
  4. Repulsive

Explanation

Each wire produces a magnetic field that acts on the other. Applying the right-hand rule shows that parallel currents in the same direction experience an attractive force, while antiparallel currents repel. The force per unit length is μ₀I₁I₂/(2πd), where d is the separation.

Question 37Magnetic Effects of Current and Magnetism

A galvanometer of resistance G is converted into an ammeter of range I using a shunt resistance S connected in parallel. Which relation correctly gives the required shunt?

  1. S = I G / I_g
  2. S = (I − I_g) G / I_g
  3. S = I_g G / (I − I_g) Correct answer.
  4. S = I_g I / G

Explanation

The shunt carries the excess current I − I_g while the galvanometer carries its full-scale current I_g, and the two are at the same potential difference. Equating I_g G = (I − I_g) S gives S = I_g G/(I − I_g). A small shunt resistance diverts most of the current, which is why an ammeter has low resistance.

Question 38Magnetic Effects of Current and Magnetism

A moving-coil galvanometer is converted into a voltmeter of range V by connecting a resistance R in series with it. If the galvanometer has resistance G and full-scale current I_g, then R equals:

  1. V/I_g − G Correct answer.
  2. V G/I_g
  3. I_g G/V
  4. V/I_g + G

Explanation

The series resistance limits the current to I_g when the full voltage V is applied across the combination. Thus V = I_g(G + R), which rearranges to R = V/I_g − G. The high series resistance is why a voltmeter draws very little current from the circuit.

Question 39Magnetic Effects of Current and Magnetism

A charged particle enters a uniform magnetic field with its velocity perpendicular to the field. The path followed by the particle is:

  1. A helix of constant pitch
  2. A straight line along the field
  3. A circle whose radius is mv/(qB) Correct answer.
  4. A parabola with increasing speed

Explanation

The magnetic force qvB acts perpendicular to the velocity, providing a centripetal force. Setting qvB = mv²/r gives r = mv/(qB), so the particle moves in a circle at constant speed. The field does no work, so the kinetic energy stays constant.

Question 40Magnetic Effects of Current and Magnetism

A magnetic dipole of moment m is placed in a uniform magnetic field B. The torque acting on it is given by:

  1. τ = m · B
  2. τ = m × B Correct answer.
  3. τ = m/B
  4. τ = mB²

Explanation

The torque on a magnetic dipole is the cross product τ = m × B, with magnitude mB sinθ. The torque is zero when the dipole is aligned with the field and maximum when it is perpendicular. This torque tends to align the dipole moment with the field, as in a compass needle.

Question 41Magnetic Effects of Current and Magnetism

A bar magnet is placed with its axis along the magnetic meridian and its north pole pointing north. The neutral point on its axis lies at a distance where the field of the magnet equals the horizontal component of Earth's field. If the magnet's field on the axis at distance d is B_axial, the neutral point occurs where:

  1. B_axial is zero at every point on the axis
  2. B_axial equals B_H and the two fields are in the same direction
  3. B_axial equals B_H and the two fields are oppositely directed Correct answer.
  4. B_axial is twice B_H in the same direction

Explanation

A neutral point is a location where the resultant magnetic field is zero, which requires the magnet's field and Earth's horizontal component to be equal in magnitude and opposite in direction. On the axial line of a magnet whose north pole faces north, the magnet's field points south and so opposes B_H. At that point a compass needle would set in any direction because no net field acts on it.

Question 42Magnetic Effects of Current and Magnetism

A solenoid of length L has N turns and carries current I. Treating it as long compared with its diameter, the magnetic field inside near the centre is approximately:

  1. μ₀NI/L²
  2. μ₀NI/L Correct answer.
  3. μ₀I/(2πL)
  4. μ₀NI/(2L)

Explanation

Applying Ampère's law to a rectangular loop that encloses part of the solenoid winding gives B L = μ₀ N I, so B = μ₀NI/L. The field inside a long solenoid is nearly uniform and parallel to the axis, and it is weak outside. Inserting a soft iron core multiplies the field by the relative permeability.

Question 43Magnetic Effects of Current and Magnetism

A circular coil of N turns and area A carries current I. Its magnetic dipole moment has magnitude:

  1. NIA Correct answer.
  2. NI/A
  3. IA/N
  4. NIA²

Explanation

Each turn behaves as a magnetic dipole of moment IA, and with N turns the moments add to give m = NIA. The direction of the moment is along the axis of the coil, given by the right-hand rule. This is why a current loop behaves like a small bar magnet.

Question 44Magnetic Effects of Current and Magnetism

A galvanometer coil of 50 Ω resistance shows full-scale deflection for a current of 2 mA. What shunt resistance converts it into an ammeter reading up to 1 A?

  1. About 25.0 Ω
  2. About 0.050 Ω
  3. About 0.100 Ω Correct answer.
  4. About 2.00 Ω

Explanation

Using S = I_g G/(I − I_g) with I_g = 0.002 A, G = 50 Ω and I = 1 A gives S = (0.002 × 50)/(0.998) ≈ 0.100 Ω. The shunt is very small compared with the galvanometer resistance, so almost all the current bypasses the coil. This is why an ammeter must have low resistance.

Question 45Magnetic Effects of Current and Magnetism

A proton and an alpha particle enter a uniform magnetic field with the same speed and perpendicular to the field. The ratio of the radius of the proton's path to that of the alpha particle is:

  1. 1 : 4
  2. 1 : 1
  3. 1 : 2 Correct answer.
  4. 2 : 1

Explanation

The radius is r = mv/(qB), so for equal speeds r is proportional to m/q. The alpha particle has about four times the proton mass and twice the charge, giving m/q twice that of the proton. Hence r_p : r_α = 1 : 2, and the alpha particle curves less sharply.

Question 46Magnetic Effects of Current and Magnetism

A charged particle moves in a region of crossed electric and magnetic fields with velocity v = E/B, the two fields being mutually perpendicular and v perpendicular to both. In this situation the particle:

  1. Follows a circular path because only the magnetic force acts
  2. Is deflected toward the electric field direction because the magnetic force vanishes
  3. Accelerates along the magnetic field direction
  4. Passes undeviated because the electric and magnetic forces balance Correct answer.

Explanation

When v = E/B, the electric force qE and the magnetic force qvB are equal in magnitude and opposite in direction, so the net force is zero. The particle therefore travels in a straight line at constant speed. This velocity-selection condition is the basis of the Thomson experiment and of velocity selectors in mass spectrometers.

Question 47Electromagnetic Induction and Alternating Currents

A bar magnet is pushed towards a stationary closed coil. According to Lenz's law, the induced current in the coil produces a magnetic field that

  1. alternates direction continuously as the magnet moves
  2. assists the motion of the magnet approaching the coil
  3. is directed parallel to the axis of the coil with no effect on motion
  4. opposes the motion of the magnet approaching the coil Correct answer.

Explanation

Lenz's law states that the induced current flows in a direction that opposes the change in magnetic flux producing it. When the magnet approaches, the flux through the coil increases, so the induced field repels the approaching pole and thereby opposes the motion. This is a direct consequence of conservation of energy, since assisting the motion would create energy without external work.

Question 48Electromagnetic Induction and Alternating Currents

Faraday's law of electromagnetic induction in the form e = -dΦ/dt tells us that the magnitude of the induced emf in a circuit equals

  1. the square of the flux linked with the circuit
  2. the product of flux and the resistance of the circuit
  3. the rate of change of magnetic flux linked with the circuit Correct answer.
  4. the total magnetic flux linked with the circuit

Explanation

The time derivative dΦ/dt represents how quickly the magnetic flux linkage changes, and the induced emf is proportional to this rate. A large but steady flux produces no emf at all, because its rate of change is zero. The minus sign merely expresses Lenz's law, giving the direction of the induced emf.

Question 49Electromagnetic Induction and Alternating Currents

The SI unit of self-inductance, the henry, is equivalent to which combination of base SI units?

  1. weber per second
  2. volt per ampere second
  3. volt second per ampere Correct answer.
  4. joule per ampere

Explanation

Self-inductance is defined by e = -L dI/dt, so L has units of volt divided by ampere per second, that is, volt second per ampere. Equivalently, one henry is one weber per ampere, since a weber is a volt second. This unit combination is what the henry represents in the SI system.

Question 50Electromagnetic Induction and Alternating Currents

In a purely capacitive AC circuit, the current relative to the applied voltage is

  1. leading the voltage by a quarter cycle Correct answer.
  2. in phase with the voltage
  3. lagging the voltage by half a cycle
  4. lagging the voltage by a quarter cycle

Explanation

For a capacitor the charge follows the voltage, so the current, being the rate of change of charge, reaches its peak before the voltage does. The current therefore leads the applied voltage by 90 degrees, a quarter of a cycle. This phase lead is the basis of the capacitive reactance Xc = 1/(ωC).