301. Ampere’s law is applied to a closed path around a long straight wire, but the path is not circular and the distance from the wire varies along the path. The safest conclusion is that
ⓐ. Ampere’s law applies, but \(B\) is not constant
ⓑ. Ampere’s law becomes false on a non-circular path
ⓒ. \(\vec{B}=0\) on any path without symmetry
ⓓ. enclosed current must be zero for every such path
Correct Answer: Ampere’s law applies, but \(B\) is not constant
Explanation: Ampere’s circuital law is valid for any closed path. However, using it to find \(B\) easily requires a path where symmetry makes the magnetic field simple along the path. For a long straight wire, a circular path centred on the wire keeps \(B\) constant in magnitude and tangent to the path. A non-circular path may still enclose the current, so the line integral equals \(\mu_0I_{\text{enc}}\). The difficulty is not the validity of the law, but the inability to replace the integral by \(B\) times a simple circumference.
302. In the Ampere-law derivation of the magnetic field inside a long solenoid, a rectangular Amperian loop is chosen with one long side inside the solenoid and one long side outside. The outside contribution is neglected because
ⓐ. outside field is nearly zero
ⓑ. outside current is zero
ⓒ. outside field lines are open
ⓓ. inside field is transverse
Correct Answer: outside field is nearly zero
Explanation: A long ideal solenoid has a nearly uniform magnetic field inside and a very weak external field. In the rectangular Amperian loop, the long side inside contributes approximately \(Bl\). The long side outside contributes nearly zero because the external field is treated as negligible. The short sides contribute zero because their path elements are perpendicular to the inside magnetic field direction. This reasoning gives \(Bl=\mu_0(nl)I\), so \(B=\mu_0nI\). The derivation depends on the long-solenoid approximation and symmetry.
303. Study the region description for an ideal toroid with inner radius \(a\), outer radius \(b\), total turns \(N\), and current \(I\).
| Region | Position of Amperian circle | Ideal magnetic field |
|---|
| P | \(r\lt a\) | Approximately zero |
| Q | \(a\lt r\lt b\) | \(\frac{\mu_0NI}{2\pi r}\) |
| R | \(r\gt b\) | Approximately zero |
The reason region Q has a non-zero field is that the Amperian circle there
ⓐ. has no current linkage at all
ⓑ. links the \(N\) current turns of the toroid
ⓒ. is outside the toroidal winding completely
ⓓ. makes \(\mu_0\) vanish in the core
Correct Answer: links the \(N\) current turns of the toroid
Explanation: In an ideal toroid, the magnetic field is mainly confined to the wound core region. For an Amperian circle with \(a\lt r\lt b\), the loop links the toroidal windings, so the enclosed current linkage is \(NI\). Ampere’s law gives \(B(2\pi r)=\mu_0NI\), leading to \(B=\frac{\mu_0NI}{2\pi r}\). In the central hollow region and outside the toroid, the net current linkage is treated as zero in the ideal model. The region-based distinction is essential because the same formula is not used everywhere.
304. Two long parallel wires carry currents \(3.0\,\text{A}\) and \(6.0\,\text{A}\) in the same direction and are separated by \(0.30\,\text{m}\). The point where their magnetic fields cancel lies between the wires at a distance from the \(3.0\,\text{A}\) wire equal to
ⓐ. \(0.15\,\text{m}\)
ⓑ. \(0.20\,\text{m}\)
ⓒ. \(0.30\,\text{m}\)
ⓓ. \(0.10\,\text{m}\)
Correct Answer: \(0.10\,\text{m}\)
Explanation: \( \textbf{Field of a long straight wire:} \)
\[
B=\frac{\mu_0I}{2\pi r}
\]
\( \textbf{Same-direction currents:} \) Between the wires, the magnetic fields due to the two wires are opposite in direction.
\( \textbf{Let the neutral point be \(x\) from the \(3.0\,\text{A}\) wire:} \)
\[
\frac{\mu_0(3.0)}{2\pi x}=\frac{\mu_0(6.0)}{2\pi(0.30-x)}
\]
\( \textbf{Cancel common factors:} \)
\[
\frac{3.0}{x}=\frac{6.0}{0.30-x}
\]
\( \textbf{Cross multiply:} \)
\[
3.0(0.30-x)=6.0x
\]
\[
0.90-3.0x=6.0x
\]
\[
0.90=9.0x
\]
\[
x=0.10\,\text{m}
\]
The neutral point is closer to the smaller current because the smaller current needs a smaller distance to produce the same field.
\( \textbf{Final answer:} \) The point is \(0.10\,\text{m}\) from the \(3.0\,\text{A}\) wire.
305. Two identical long wires are perpendicular to the page and pass through points P and Q. The currents in both wires are out of the page. At the midpoint between P and Q, the magnetic fields due to the two wires are
ⓐ. equal in magnitude and in the same direction
ⓑ. unequal in magnitude and opposite in direction
ⓒ. zero individually because the midpoint lies between wires
ⓓ. equal in magnitude and opposite in direction
Correct Answer: equal in magnitude and opposite in direction
Explanation: The midpoint is at the same distance from both identical wires, so the magnetic field magnitudes there are equal. For current out of the page, field lines are anticlockwise around each wire. At the midpoint, the tangent direction due to one wire is opposite to the tangent direction due to the other. Therefore, the two fields cancel at the midpoint. This cancellation occurs for equal same-direction currents at the midpoint, not because each individual field is zero.
306. A finite straight wire carries current \(10\,\text{A}\). At a point a perpendicular distance \(5.0\,\text{cm}\) from the wire, the two end angles in \(B=\frac{\mu_0I}{4\pi r}(\sin\phi_1+\sin\phi_2)\) are both \(45^\circ\). Taking \(\frac{\mu_0}{4\pi}=1.0\times10^{-7}\,\text{T m A}^{-1}\), the magnetic field is closest to
ⓐ. \(1.4\times10^{-5}\,\text{T}\)
ⓑ. \(2.8\times10^{-5}\,\text{T}\)
ⓒ. \(4.0\times10^{-5}\,\text{T}\)
ⓓ. \(8.0\times10^{-5}\,\text{T}\)
Correct Answer: \(2.8\times10^{-5}\,\text{T}\)
Explanation: \( \textbf{Given data:} \) \(I=10\,\text{A}\), \(r=5.0\,\text{cm}=5.0\times10^{-2}\,\text{m}\), and \(\phi_1=\phi_2=45^\circ\).
\( \textbf{Finite-wire formula:} \)
\[
B=\frac{\mu_0I}{4\pi r}(\sin\phi_1+\sin\phi_2)
\]
\( \textbf{Angle term:} \)
\[
\sin45^\circ+\sin45^\circ=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}
\]
\( \textbf{Substitution:} \)
\[
B=(1.0\times10^{-7})\frac{10}{5.0\times10^{-2}}(\sqrt{2})
\]
\( \textbf{Current-distance factor:} \)
\[
\frac{10}{5.0\times10^{-2}}=200
\]
\( \textbf{Field value:} \)
\[
B=(1.0\times10^{-7})(200)(1.414)
\]
\[
B\approx2.8\times10^{-5}\,\text{T}
\]
The field is less than the long-wire value for the same \(I\) and \(r\) because the wire is finite.
\( \textbf{Final answer:} \) \(B\approx2.8\times10^{-5}\,\text{T}\).
307. For a circular loop carrying current \(I\), let \(B_0\) be the magnetic field at its centre. At an axial point where \(x=\sqrt{3}R\), the magnetic field is
ⓐ. \(\frac{B_0}{2}\)
ⓑ. \(\frac{B_0}{4}\)
ⓒ. \(\frac{B_0}{8}\)
ⓓ. \(\frac{B_0}{16}\)
Correct Answer: \(\frac{B_0}{8}\)
Explanation: \( \textbf{Axial field of a circular loop:} \)
\[
B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}
\]
\( \textbf{At the centre:} \)
\[
B_0=\frac{\mu_0 I}{2R}
\]
\( \textbf{Given axial distance:} \)
\[
x=\sqrt{3}R
\]
\[
R^2+x^2=R^2+3R^2=4R^2
\]
\( \textbf{Denominator power:} \)
\[
(R^2+x^2)^{3/2}=(4R^2)^{3/2}=8R^3
\]
\( \textbf{Axial field:} \)
\[
B=\frac{\mu_0IR^2}{2(8R^3)}
\]
\[
B=\frac{\mu_0 I}{16R}
\]
\( \textbf{Compare with \(B_0\):} \)
\[
B_0=\frac{\mu_0I}{2R}
\]
\[
B=\frac{B_0}{8}
\]
The field drops quickly along the axis because the denominator contains the power \(\frac{3}{2}\).
\( \textbf{Final answer:} \) The field is \(\frac{B_0}{8}\).
308. A circular current loop is viewed from one side, and the current appears anticlockwise. The face seen by the observer behaves like
ⓐ. a south pole face
ⓑ. a neutral face with no magnetic polarity
ⓒ. a north pole face
ⓓ. an electric positive plate
Correct Answer: a north pole face
Explanation: Curl the fingers of the right hand in the direction of current in the loop. If the current appears anticlockwise to the observer, the thumb points toward the observer. The magnetic field therefore emerges from that face of the loop. A face from which magnetic field lines emerge is treated as a north pole face in the bar-magnet analogy. This is a polarity description of the loop, while the actual magnetic field lines remain closed loops.
309. A current loop is compared with a bar magnet. The best statement is that the loop
ⓐ. it contains isolated magnetic north charge
ⓑ. electric field lines begin from its north face
ⓒ. it has zero magnetic moment when current flows
ⓓ. it has magnetic dipole moment like a small magnet
Correct Answer: it has magnetic dipole moment like a small magnet
Explanation: A current loop has magnetic dipole moment \(\vec{m}=I\vec{A}\) for one turn and \(\vec{m}=NI\vec{A}\) for \(N\) turns. Because of this magnetic moment, it behaves like a small magnetic dipole in an external magnetic field. It can experience torque and has a preferred alignment with the field. The loop analogy with a bar magnet does not mean there are isolated magnetic charges at its faces. Magnetic field lines remain closed, so the dipole picture is about field pattern and torque behaviour.
310. A magnetic dipole is slightly displaced from its stable equilibrium in a uniform magnetic field. The torque that appears tends to
ⓐ. turn \(\vec{m}\) toward \(\vec{B}\)
ⓑ. increase its potential energy
ⓒ. keep \(\vec{m}\) perpendicular to \(\vec{B}\)
ⓓ. make the magnetic field vanish
Correct Answer: turn \(\vec{m}\) toward \(\vec{B}\)
Explanation: Stable equilibrium occurs when \(\vec{m}\) is parallel to \(\vec{B}\). At that position, the potential energy \(U=-mB\cos\theta\) is minimum. If the dipole is slightly displaced, the torque \(\vec{\tau}=\vec{m}\times\vec{B}\) tends to reduce the angle \(\theta\). Reducing \(\theta\) brings the dipole back toward the lower-energy aligned position. This is why the parallel orientation is stable, while the anti-parallel orientation is unstable even though torque is zero there.
311. A rectangular loop lies in the plane of the page and carries clockwise current. A uniform magnetic field is directed to the right in the plane of the page. The magnetic forces on the upper and lower horizontal sides are
ⓐ. out of the page on both sides
ⓑ. into the page on both sides
ⓒ. zero on both sides
ⓓ. equal and opposite along the magnetic field
Correct Answer: zero on both sides
Explanation: The force on a straight current-carrying side is \(F=IlB\sin\theta\). The upper and lower sides of the rectangular loop are horizontal, so their currents are along or opposite to the right-left direction. The magnetic field is also horizontal to the right. Therefore, the angle between each horizontal side’s current direction and \(\vec{B}\) is either \(0^\circ\) or \(180^\circ\). In both cases, \(\sin\theta=0\), so the force on those sides is zero. The vertical sides can still experience forces because their current directions are perpendicular to the field.
312. A long straight wire carries current upward. A positive charge at a point east of the wire moves upward, parallel to the wire. The magnetic force on the charge is directed
ⓐ. east
ⓑ. west
ⓒ. upward
ⓓ. downward
Correct Answer: west
Explanation: First find the magnetic field due to the wire at the point east of it. With current upward, the right-hand thumb rule gives magnetic field toward north at the east-side point. The positive charge moves upward, so \(\vec{v}\) is upward and \(\vec{B}\) is north. Taking east as \(+x\), north as \(+y\), and upward as \(+z\), the force direction is \(\vec{v}\times\vec{B}=+z\times+y=-x\). The negative \(x\)-direction is west. The problem combines field direction due to a current with force direction on a moving charge.
313. A long straight wire carries current \(10\,\text{A}\). At a point \(2.0\,\text{cm}\) from the wire, a particle of charge magnitude \(1.6\times10^{-19}\,\text{C}\) moves perpendicular to the wire’s magnetic field with speed \(2.0\times10^6\,\text{m s}^{-1}\). Taking \(\mu_0=4\pi\times10^{-7}\,\text{T m A}^{-1}\), the magnetic force magnitude is
ⓐ. \(3.2\times10^{-17}\,\text{N}\)
ⓑ. \(1.6\times10^{-17}\,\text{N}\)
ⓒ. \(6.4\times10^{-17}\,\text{N}\)
ⓓ. \(1.6\times10^{-16}\,\text{N}\)
Correct Answer: \(3.2\times10^{-17}\,\text{N}\)
Explanation: \( \textbf{Field due to long wire:} \)
\[
B=\frac{\mu_0I}{2\pi r}
\]
\( \textbf{Given data:} \) \(I=10\,\text{A}\), \(r=2.0\,\text{cm}=2.0\times10^{-2}\,\text{m}\).
\( \textbf{Substitute for \(B\):} \)
\[
B=\frac{(4\pi\times10^{-7})(10)}{2\pi(2.0\times10^{-2})}
\]
\( \textbf{Simplify:} \)
\[
B=\frac{4.0\times10^{-6}}{4.0\times10^{-2}}=1.0\times10^{-4}\,\text{T}
\]
\( \textbf{Force on moving charge:} \)
\[
F=|q|vB\sin90^\circ
\]
\( \textbf{Substitution:} \)
\[
F=(1.6\times10^{-19})(2.0\times10^6)(1.0\times10^{-4})
\]
\[
F=3.2\times10^{-17}\,\text{N}
\]
This is a two-step magnetic interaction: current first produces \(\vec{B}\), then \(\vec{B}\) produces force on a moving charge.
\( \textbf{Final answer:} \) The magnetic force is \(3.2\times10^{-17}\,\text{N}\).
314. In a moving-coil galvanometer, the soft iron core is used mainly to
ⓐ. stop current from flowing through the coil
ⓑ. make the magnetic field radial and stronger near the coil
ⓒ. convert the galvanometer directly into a voltmeter without resistance
ⓓ. make the restoring torque independent of deflection
Correct Answer: make the magnetic field radial and stronger near the coil
Explanation: The soft iron core in a moving-coil galvanometer helps shape the magnetic field into a radial field. A radial field keeps the plane of the coil parallel to the field during rotation, so the area vector remains perpendicular to \(\vec{B}\). This keeps the magnetic torque proportional to current for all deflections. The soft iron core also increases the magnetic field strength because of its high permeability. It does not replace the need for shunt or series resistance in meter conversion.
315. A galvanometer coil has \(N=200\), \(A=1.5\times10^{-4}\,\text{m}^2\), \(B=0.40\,\text{T}\), and \(k=2.4\times10^{-5}\,\text{N m rad}^{-1}\). The current required for a deflection of \(0.10\,\text{rad}\) is
ⓐ. \(1.0\times10^{-4}\,\text{A}\)
ⓑ. \(4.0\times10^{-4}\,\text{A}\)
ⓒ. \(2.0\times10^{-4}\,\text{A}\)
ⓓ. \(8.0\times10^{-4}\,\text{A}\)
Correct Answer: \(2.0\times10^{-4}\,\text{A}\)
Explanation: \( \textbf{Torque balance in radial field:} \)
\[
NIAB=k\phi
\]
\( \textbf{Required current:} \)
\[
I=\frac{k\phi}{NAB}
\]
\( \textbf{Given data:} \) \(N=200\), \(A=1.5\times10^{-4}\,\text{m}^2\), \(B=0.40\,\text{T}\), \(k=2.4\times10^{-5}\,\text{N m rad}^{-1}\), and \(\phi=0.10\,\text{rad}\).
\( \textbf{Substitution:} \)
\[
I=\frac{(2.4\times10^{-5})(0.10)}{(200)(1.5\times10^{-4})(0.40)}
\]
\( \textbf{Numerator:} \)
\[
(2.4\times10^{-5})(0.10)=2.4\times10^{-6}
\]
\( \textbf{Denominator:} \)
\[
(200)(1.5\times10^{-4})=3.0\times10^{-2}
\]
\[
(3.0\times10^{-2})(0.40)=1.2\times10^{-2}
\]
\( \textbf{Current:} \)
\[
I=\frac{2.4\times10^{-6}}{1.2\times10^{-2}}=2.0\times10^{-4}\,\text{A}
\]
The radial field keeps the torque proportional to current, so the balance relation can be used directly for the required deflection.
\( \textbf{Final answer:} \) The required current is \(2.0\times10^{-4}\,\text{A}\).
316. A galvanometer coil has \(N=100\), \(A=2.0\times10^{-4}\,\text{m}^2\), \(B=0.50\,\text{T}\), and \(k=1.0\times10^{-6}\,\text{N m rad}^{-1}\). The current required for a deflection of \(0.20\,\text{rad}\) is
ⓐ. \(2.0\times10^{-5}\,\text{A}\)
ⓑ. \(1.0\times10^{-5}\,\text{A}\)
ⓒ. \(4.0\times10^{-5}\,\text{A}\)
ⓓ. \(1.0\times10^{-4}\,\text{A}\)
Correct Answer: \(2.0\times10^{-5}\,\text{A}\)
Explanation: \( \textbf{Equilibrium condition:} \)
\[
NIAB=k\phi
\]
\( \textbf{Solve for current:} \)
\[
I=\frac{k\phi}{NAB}
\]
\( \textbf{Given data:} \) \(N=100\), \(A=2.0\times10^{-4}\,\text{m}^2\), \(B=0.50\,\text{T}\), \(k=1.0\times10^{-6}\,\text{N m rad}^{-1}\), and \(\phi=0.20\,\text{rad}\).
\( \textbf{Substitution:} \)
\[
I=\frac{(1.0\times10^{-6})(0.20)}{(100)(2.0\times10^{-4})(0.50)}
\]
\( \textbf{Numerator:} \)
\[
(1.0\times10^{-6})(0.20)=2.0\times10^{-7}
\]
\( \textbf{Denominator:} \)
\[
(100)(2.0\times10^{-4})=2.0\times10^{-2}
\]
\[
(2.0\times10^{-2})(0.50)=1.0\times10^{-2}
\]
\( \textbf{Current:} \)
\[
I=\frac{2.0\times10^{-7}}{1.0\times10^{-2}}=2.0\times10^{-5}\,\text{A}
\]
A smaller torsional constant or larger \(NAB\) would reduce the current needed for the same deflection.
\( \textbf{Final answer:} \) The required current is \(2.0\times10^{-5}\,\text{A}\).
317. A converted voltmeter has resistance \(2000\,\Omega\) and is connected across a \(2000\,\Omega\) resistor in a circuit. The measured branch resistance becomes
ⓐ. \(2000\,\Omega\)
ⓑ. \(4000\,\Omega\)
ⓒ. \(1000\,\Omega\)
ⓓ. \(0\,\Omega\)
Correct Answer: \(1000\,\Omega\)
Explanation: \( \textbf{Connection type:} \) A voltmeter is connected in parallel with the resistor being measured.
\( \textbf{Given resistances:} \)
\[
R=2000\,\Omega,\qquad R_V=2000\,\Omega
\]
\( \textbf{Parallel equivalent:} \)
\[
R_{\text{eq}}=\frac{RR_V}{R+R_V}
\]
\( \textbf{Substitution:} \)
\[
R_{\text{eq}}=\frac{(2000)(2000)}{2000+2000}
\]
\[
R_{\text{eq}}=\frac{4.0\times10^6}{4000}
\]
\[
R_{\text{eq}}=1000\,\Omega
\]
The voltmeter resistance is not high enough here, so it significantly loads the circuit.
\( \textbf{Final answer:} \) The branch resistance becomes \(1000\,\Omega\).
318. Match the direction rule with its most suitable use.
| Rule | Use |
|---|
| P. Right-hand rule for \(\vec{v}\times\vec{B}\) | 1. Magnetic field direction around a straight current-carrying wire |
| Q. Right-hand thumb rule for a straight wire | 2. Magnetic force direction on a positive moving charge |
| R. Right-hand grip rule for a current loop | 3. Direction of magnetic moment of a current loop |
| S. Fleming’s left-hand rule | 4. Force direction on a current-carrying conductor in a magnetic field |
ⓐ. P-1, Q-2, R-3, S-4
ⓑ. P-2, Q-3, R-1, S-4
ⓒ. P-4, Q-1, R-2, S-3
ⓓ. P-2, Q-1, R-3, S-4
Correct Answer: P-2, Q-1, R-3, S-4
Explanation: The right-hand rule for \(\vec{v}\times\vec{B}\) gives the magnetic-force direction for a positive moving charge. The right-hand thumb rule for a straight wire gives the circular magnetic field direction around the wire when the thumb points along current. The right-hand grip rule for a loop gives the direction of the loop’s magnetic moment or axial field. Fleming’s left-hand rule is commonly used for force direction on a current-carrying conductor in a magnetic field. Keeping the rules separate prevents mixing up field direction with force direction.
319. A data record for a particle moving perpendicular to a magnetic field shows that doubling \(B\) halves the radius and doubles the magnetic force. This is consistent because
ⓐ. both \(r\) and \(F\) are proportional to \(B\)
ⓑ. \(r\propto1/B\) while \(F\propto B\)
ⓒ. both \(r\) and \(F\) are proportional to \(1/B\)
ⓓ. radius changes only if speed changes
Correct Answer: \(r\propto1/B\) while \(F\propto B\)
Explanation: For perpendicular magnetic motion, the force magnitude is \(F=|q|vB\). With \(|q|\) and \(v\) fixed, the magnetic force is directly proportional to \(B\). The circular radius is \(r=\frac{mv}{|q|B}\). With \(m\), \(|q|\), and \(v\) fixed, radius is inversely proportional to \(B\). Therefore, doubling \(B\) doubles \(F\) but halves \(r\), so the two observations are not contradictory.
320. A proposed solution says: “A closed current loop in a uniform magnetic field has zero net force, so it cannot have torque.” The error in this claim is that
ⓐ. a closed loop always has non-zero net force in a uniform field
ⓑ. torque requires net force to be non-zero in every situation
ⓒ. magnetic forces on the sides of a loop are always zero
ⓓ. zero net force does not rule out a couple producing torque
Correct Answer: zero net force does not rule out a couple producing torque
Explanation: In a uniform magnetic field, the net magnetic force on a closed current loop is zero. However, forces on different sides of the loop may be equal and opposite but act along different lines of action. Such a pair can form a couple and produce torque. The loop may rotate even though it does not translate. This distinction between net force and torque is central to current-loop and galvanometer behaviour.
