301. A beam passes undeflected through crossed fields with \(E=5.0\times10^4\,\text{V m}^{-1}\) and \(B=0.25\,\text{T}\). The selected particles then enter a second uniform magnetic field of \(0.40\,\text{T}\) perpendicular to their velocity. If \(\frac{m}{|q|}=2.0\times10^{-7}\,\text{kg C}^{-1}\), what is the radius of their circular path?
ⓐ. \(0.05\,\text{m}\)
ⓑ. \(0.20\,\text{m}\)
ⓒ. \(0.10\,\text{m}\)
ⓓ. \(0.40\,\text{m}\)
Correct Answer: \(0.10\,\text{m}\)
Explanation: \( \textbf{Velocity selector relation:} \)
\[
v=\frac{E}{B}
\]
\( \textbf{Selected speed:} \)
\[
v=\frac{5.0\times10^4}{0.25}=2.0\times10^5\,\text{m s}^{-1}
\]
\( \textbf{Magnetic radius relation:} \)
\[
r=\frac{mv}{|q|B'}
\]
\( \textbf{Use mass-to-charge ratio:} \)
\[
r=\left(\frac{m}{|q|}\right)\frac{v}{B'}
\]
\( \textbf{Substitution:} \)
\[
r=(2.0\times10^{-7})\frac{2.0\times10^5}{0.40}
\]
\( \textbf{Numerator simplification:} \)
\[
(2.0\times10^{-7})(2.0\times10^5)=4.0\times10^{-2}
\]
\( \textbf{Final calculation:} \)
\[
r=\frac{4.0\times10^{-2}}{0.40}=1.0\times10^{-1}\,\text{m}
\]
\( \textbf{Final answer:} \) The radius is \(0.10\,\text{m}\).
302. A long straight wire carries current \(10\,\text{A}\). A second wire of length \(0.30\,\text{m}\) carrying current \(5.0\,\text{A}\) is placed parallel to it at a distance \(0.15\,\text{m}\). What is the magnitude of the magnetic force on this \(0.30\,\text{m}\) length? Use \(\mu_0=4\pi\times10^{-7}\,\text{T m A}^{-1}\).
ⓐ. \(2.0\times10^{-5}\,\text{N}\)
ⓑ. \(1.0\times10^{-5}\,\text{N}\)
ⓒ. \(4.0\times10^{-5}\,\text{N}\)
ⓓ. \(8.0\times10^{-5}\,\text{N}\)
Correct Answer: \(2.0\times10^{-5}\,\text{N}\)
Explanation: \( \textbf{Field due to first wire:} \)
\[
B_1=\frac{\mu_0I_1}{2\pi d}
\]
\( \textbf{Force on second wire:} \)
\[
F=I_2lB_1
\]
\( \textbf{Combined relation:} \)
\[
F=\frac{\mu_0I_1I_2l}{2\pi d}
\]
\( \textbf{Substitution:} \)
\[
F=\frac{(4\pi\times10^{-7})(10)(5.0)(0.30)}{2\pi(0.15)}
\]
\( \textbf{Current-length product:} \)
\[
(10)(5.0)(0.30)=15
\]
\( \textbf{Simplification:} \)
\[
F=\frac{60\pi\times10^{-7}}{0.30\pi}=200\times10^{-7}\,\text{N}
\]
\[
F=2.0\times10^{-5}\,\text{N}
\]
\( \textbf{Direction note:} \) Same-direction currents attract and opposite-direction currents repel, but the magnitude above depends on current magnitudes and separation.
\( \textbf{Final answer:} \) The force magnitude is \(2.0\times10^{-5}\,\text{N}\).
303. A galvanometer has \(G=40\,\Omega\) and \(I_g=2.0\,\text{mA}\). Which added resistance converts it into a \(4.0\,\text{V}\) voltmeter?
ⓐ. \(1960\,\Omega\) in parallel
ⓑ. \(1960\,\Omega\) in series
ⓒ. \(0.020\,\Omega\) in series
ⓓ. \(0.020\,\Omega\) in parallel
Correct Answer: \(1960\,\Omega\) in series
Explanation: \( \textbf{Given:} \) \(G=40\,\Omega\), \(I_g=2.0\,\text{mA}=2.0\times10^{-3}\,\text{A}\), and \(V=4.0\,\text{V}\).
\( \textbf{Voltmeter conversion rule:} \) A high resistance must be connected in series with the galvanometer.
\( \textbf{Series resistance formula:} \)
\[
R=\frac{V}{I_g}-G
\]
\( \textbf{Total resistance needed:} \)
\[
\frac{V}{I_g}=\frac{4.0}{2.0\times10^{-3}}=2000\,\Omega
\]
\( \textbf{Subtract galvanometer resistance:} \)
\[
R=2000-40=1960\,\Omega
\]
\( \textbf{Connection check:} \) The resistance is connected in series because a voltmeter must have high total resistance.
\( \textbf{Final answer:} \) The required resistance is \(1960\,\Omega\) in series.
304. An Amperian loop encloses currents \(4\,\text{A}\) out of the page, \(7\,\text{A}\) into the page, \(5\,\text{A}\) out of the page, and \(2\,\text{A}\) outside the loop. Taking out of the page as positive, what is \(\oint\vec{B}\cdot d\vec{l}\)?
ⓐ. \(\mu_0(4\,\text{A})\)
ⓑ. \(\mu_0(8\,\text{A})\)
ⓒ. \(\mu_0(2\,\text{A})\)
ⓓ. \(\mu_0(18\,\text{A})\)
Correct Answer: \(\mu_0(2\,\text{A})\)
Explanation: \( \textbf{Ampere's circuital law:} \)
\[
\oint\vec{B}\cdot d\vec{l}=\mu_0I_{\text{enc}}
\]
\( \textbf{Enclosed-current rule:} \) Only currents passing through the surface bounded by the loop are counted.
\( \textbf{Sign convention:} \) Out of the page is positive, and into the page is negative.
\( \textbf{Algebraic enclosed current:} \)
\[
I_{\text{enc}}=4-7+5=2\,\text{A}
\]
\( \textbf{Outside current:} \) The \(2\,\text{A}\) current outside the loop is not included in \(I_{\text{enc}}\).
\( \textbf{Line integral:} \)
\[
\oint\vec{B}\cdot d\vec{l}=\mu_0(2\,\text{A})
\]
\( \textbf{Final answer:} \) The line integral is \(\mu_0(2\,\text{A})\).
305. A particle enters a uniform magnetic field with speed \(5.0\times10^5\,\text{m s}^{-1}\) at an angle \(53^\circ\) with the field. Take \(\sin53^\circ=0.8\) and \(\cos53^\circ=0.6\). If \(m=4.0\times10^{-26}\,\text{kg}\), \(|q|=2.0\times10^{-19}\,\text{C}\), and \(B=0.50\,\text{T}\), what is the radius of the helical path?
ⓐ. \(0.08\,\text{m}\)
ⓑ. \(0.16\,\text{m}\)
ⓒ. \(0.12\,\text{m}\)
ⓓ. \(0.20\,\text{m}\)
Correct Answer: \(0.16\,\text{m}\)
Explanation: \( \textbf{Given:} \) \(v=5.0\times10^5\,\text{m s}^{-1}\), \(\theta=53^\circ\), \(m=4.0\times10^{-26}\,\text{kg}\), \(|q|=2.0\times10^{-19}\,\text{C}\), and \(B=0.50\,\text{T}\).
\( \textbf{Perpendicular component:} \)
\[
v_\perp=v\sin\theta
\]
\( \textbf{Substitution:} \)
\[
v_\perp=(5.0\times10^5)(0.8)=4.0\times10^5\,\text{m s}^{-1}
\]
\( \textbf{Helical radius relation:} \)
\[
r=\frac{mv_\perp}{|q|B}
\]
\( \textbf{Substitution:} \)
\[
r=\frac{(4.0\times10^{-26})(4.0\times10^5)}{(2.0\times10^{-19})(0.50)}
\]
\( \textbf{Numerator:} \)
\[
(4.0\times10^{-26})(4.0\times10^5)=1.6\times10^{-20}
\]
\( \textbf{Denominator:} \)
\[
(2.0\times10^{-19})(0.50)=1.0\times10^{-19}
\]
\( \textbf{Final calculation:} \)
\[
r=\frac{1.6\times10^{-20}}{1.0\times10^{-19}}=0.16\,\text{m}
\]
\( \textbf{Final answer:} \) The radius of the helical path is \(0.16\,\text{m}\).
306. For a charged particle moving perpendicular to a uniform magnetic field, which row correctly describes the effect of doubling \(B\), keeping \(m\), \(|q|\), and \(v\) fixed?
| Row | Radius \(r\) | Time period \(T\) | Magnetic force \(F\) |
|---|
| P | Halves | Halves | Doubles |
| Q | Doubles | Doubles | Halves |
| R | Unchanged | Halves | Doubles |
| S | Halves | Unchanged | Unchanged |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: The circular radius is \(r=\frac{mv}{|q|B}\), so doubling \(B\) makes the radius half. The time period is \(T=\frac{2\pi m}{|q|B}\), so doubling \(B\) also makes the time period half. The magnetic force for perpendicular motion is \(F=|q|vB\), so doubling \(B\) doubles the force. Row P correctly combines all three effects. The important point is that a stronger magnetic field makes the path tighter and the revolution faster, even though it increases the instantaneous force magnitude.
307. A circular loop carries current clockwise as seen from the front. A uniform magnetic field is directed from left to right in the plane of the page. What is the direction of the torque tendency on the loop?
ⓐ. It tends to turn the magnetic moment from out of the page toward the left
ⓑ. It has zero torque because the field lies in the plane of the loop
ⓒ. It has zero torque because clockwise current gives no magnetic moment
ⓓ. It tends to turn the magnetic moment from into the page toward the right
Correct Answer: It tends to turn the magnetic moment from into the page toward the right
Explanation: For clockwise current as seen from the front, the magnetic moment \(\vec{m}\) points into the page by the right-hand curl rule. The magnetic field \(\vec{B}\) is directed to the right in the plane of the page. Since \(\vec{m}\) is perpendicular to \(\vec{B}\), the torque magnitude is maximum, not zero. The torque \(\vec{\tau}=\vec{m}\times\vec{B}\) acts to rotate \(\vec{m}\) toward alignment with \(\vec{B}\). Physically, the loop tends to turn so that its magnetic moment moves from into the page toward the right. The plane of the loop being parallel to \(\vec{B}\) is exactly the condition for maximum torque, not zero torque.
308. A long straight wire and an ideal toroid both have magnetic fields varying as \(\frac{1}{r}\) in their respective formulae. Which distinction is correct?
ⓐ. Wire field extends outside; ideal toroid field is core-confined
ⓑ. Both fields are radial outward from the current source
ⓒ. Both fields are zero at every point outside the conductor material
ⓓ. The wire field uses \(\mu_0\), but the toroid field never uses \(\mu_0\)
Correct Answer: Wire field extends outside; ideal toroid field is core-confined
Explanation: For a long straight wire, \(B=\frac{\mu_0I}{2\pi r}\) at distance \(r\) from the wire, and the magnetic field exists in the surrounding space. For an ideal toroid, \(B=\frac{\mu_0NI}{2\pi r}\) applies mainly inside the toroidal core. The ideal toroid's external field and field in the central empty region are taken as nearly zero. In both cases, the field direction is tangential to circular field lines, not radial outward. Both formulae contain \(\mu_0\) in free space. The similar \(\frac{1}{r}\) form should not hide the different field regions and symmetries.
309. A galvanometer has \(I_g=1.0\,\text{mA}\) and \(G=80\,\Omega\). It is converted into an ammeter using a shunt \(S=0.20\,\Omega\). What is the approximate full-scale range of the ammeter?
ⓐ. \(0.20\,\text{A}\)
ⓑ. \(0.80\,\text{A}\)
ⓒ. \(1.60\,\text{A}\)
ⓓ. \(0.40\,\text{A}\)
Correct Answer: \(0.40\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(I_g=1.0\times10^{-3}\,\text{A}\), \(G=80\,\Omega\), and \(S=0.20\,\Omega\).
\( \textbf{Parallel voltage condition:} \)
\[
I_gG=I_sS
\]
\( \textbf{Solve for shunt current:} \)
\[
I_s=\frac{I_gG}{S}
\]
\( \textbf{Substitution:} \)
\[
I_s=\frac{(1.0\times10^{-3})(80)}{0.20}
\]
\( \textbf{Simplification:} \)
\[
I_s=\frac{0.080}{0.20}=0.40\,\text{A}
\]
\( \textbf{Total current:} \)
\[
I=I_g+I_s=0.001+0.40=0.401\,\text{A}
\]
\( \textbf{Nearest range:} \) The full-scale ammeter range is approximately \(0.40\,\text{A}\).
\( \textbf{Final answer:} \) The ammeter range is about \(0.40\,\text{A}\).
310. A graph is plotted between \(\frac{F}{l}\) and \(I_1I_2\) for two long parallel wires at fixed separation \(d\). What is the slope of the graph?
ⓐ. \(\frac{\mu_0I_1I_2}{2\pi d}\)
ⓑ. \(\frac{\mu_0}{2\pi d}\)
ⓒ. \(\frac{2\pi d}{\mu_0}\)
ⓓ. \(\frac{\mu_0d}{2\pi}\)
Correct Answer: \(\frac{\mu_0}{2\pi d}\)
Explanation: \( \textbf{Force per unit length:} \)
\[
\frac{F}{l}=\frac{\mu_0I_1I_2}{2\pi d}
\]
\( \textbf{Fixed quantity:} \) The separation \(d\) is fixed.
\( \textbf{Graph variable:} \) The horizontal axis is \(I_1I_2\).
\( \textbf{Linear form:} \)
\[
\frac{F}{l}=\left(\frac{\mu_0}{2\pi d}\right)(I_1I_2)
\]
\( \textbf{Slope:} \) The slope is \(\frac{\mu_0}{2\pi d}\).
\( \textbf{Common mistake:} \) \(\frac{\mu_0I_1I_2}{2\pi d}\) is the actual force per unit length, not the slope when \(I_1I_2\) is the plotted variable.
\( \textbf{Final answer:} \) The slope is \(\frac{\mu_0}{2\pi d}\).
311. A charged particle moves in a circle of radius \(0.50\,\text{m}\) in a uniform magnetic field \(0.20\,\text{T}\). Its speed is \(2.0\times10^5\,\text{m s}^{-1}\). What is \(\frac{m}{|q|}\)?
ⓐ. \(2.5\times10^{-7}\,\text{kg C}^{-1}\)
ⓑ. \(1.0\times10^{-6}\,\text{kg C}^{-1}\)
ⓒ. \(5.0\times10^{-7}\,\text{kg C}^{-1}\)
ⓓ. \(2.0\times10^{-6}\,\text{kg C}^{-1}\)
Correct Answer: \(5.0\times10^{-7}\,\text{kg C}^{-1}\)
Explanation: \( \textbf{Given:} \) \(r=0.50\,\text{m}\), \(B=0.20\,\text{T}\), and \(v=2.0\times10^5\,\text{m s}^{-1}\).
\( \textbf{Radius relation:} \)
\[
r=\frac{mv}{|q|B}
\]
\( \textbf{Solve for mass-to-charge ratio:} \)
\[
\frac{m}{|q|}=\frac{rB}{v}
\]
\( \textbf{Substitution:} \)
\[
\frac{m}{|q|}=\frac{(0.50)(0.20)}{2.0\times10^5}
\]
\( \textbf{Numerator:} \)
\[
(0.50)(0.20)=0.100
\]
\( \textbf{Final calculation:} \)
\[
\frac{m}{|q|}=\frac{0.100}{2.0\times10^5}=5.0\times10^{-7}\,\text{kg C}^{-1}
\]
\( \textbf{Final answer:} \) \(\frac{m}{|q|}=5.0\times10^{-7}\,\text{kg C}^{-1}\).
312. Which statement best identifies a limitation of using only magnitude formulae in magnetic-force problems?
ⓐ. Magnitude formulae are always wrong for magnetic field questions
ⓑ. Direction rules are needed only for electric fields, not magnetic fields
ⓒ. Cross products are unnecessary whenever charge is negative
ⓓ. Magnitudes cannot decide attraction, repulsion, or page directions
Correct Answer: Magnitudes cannot decide attraction, repulsion, or page directions
Explanation: Magnitude formulae such as \(F=|q|vB\sin\theta\), \(F=IlB\sin\theta\), and \(B=\frac{\mu_0I}{2\pi r}\) give sizes of quantities. They do not by themselves decide whether a force is upward, downward, into the page, or out of the page. Direction questions need vector laws such as \(\vec{F}=q(\vec{v}\times\vec{B})\), \(\vec{F}=I(\vec{l}\times\vec{B})\), and right-hand rules for fields around currents. The sign of charge and the direction of current can reverse the force or field direction without changing the magnitude formula. Many mistakes occur when learners calculate the right size but assign the wrong direction.
313. A positive ion passes undeflected through crossed electric and magnetic fields with \(E=7.2\times10^4\,\text{V m}^{-1}\) and \(B=0.24\,\text{T}\). It then enters a second magnetic field of \(0.60\,\text{T}\) perpendicular to its velocity and moves in a circle of radius \(0.15\,\text{m}\). What is \(\frac{m}{q}\) for the ion?
ⓐ. \(2.0\times10^{-7}\,\text{kg C}^{-1}\)
ⓑ. \(4.0\times10^{-7}\,\text{kg C}^{-1}\)
ⓒ. \(3.0\times10^{-7}\,\text{kg C}^{-1}\)
ⓓ. \(5.0\times10^{-7}\,\text{kg C}^{-1}\)
Correct Answer: \(3.0\times10^{-7}\,\text{kg C}^{-1}\)
Explanation: \( \textbf{Velocity selector condition:} \)
\[
v=\frac{E}{B}
\]
\( \textbf{Selected speed:} \)
\[
v=\frac{7.2\times10^4}{0.24}=3.0\times10^5\,\text{m s}^{-1}
\]
\( \textbf{Circular motion in second field:} \)
\[
r=\frac{mv}{qB'}
\]
\( \textbf{Solve for mass-to-charge ratio:} \)
\[
\frac{m}{q}=\frac{rB'}{v}
\]
\( \textbf{Substitution:} \)
\[
\frac{m}{q}=\frac{(0.15)(0.60)}{3.0\times10^5}
\]
\( \textbf{Numerator:} \)
\[
(0.15)(0.60)=0.090
\]
\( \textbf{Final simplification:} \)
\[
\frac{m}{q}=\frac{0.090}{3.0\times10^5}=3.0\times10^{-7}\,\text{kg C}^{-1}
\]
\( \textbf{Final answer:} \) \(\frac{m}{q}=3.0\times10^{-7}\,\text{kg C}^{-1}\).
314. A rectangular coil has magnetic moment \(0.40\,\text{A m}^2\) and is placed in a uniform magnetic field \(0.50\,\text{T}\). The plane of the coil makes an angle \(30^\circ\) with the magnetic field. What is the torque on the coil?
ⓐ. \(0.173\,\text{N m}\)
ⓑ. \(0.10\,\text{N m}\)
ⓒ. \(0.20\,\text{N m}\)
ⓓ. \(0.346\,\text{N m}\)
Correct Answer: \(0.173\,\text{N m}\)
Explanation: \( \textbf{Given:} \) \(m=0.40\,\text{A m}^2\), \(B=0.50\,\text{T}\), and the plane of the coil makes \(30^\circ\) with \(\vec{B}\).
\( \textbf{Angle-related mistake:} \) Torque uses the angle \(\theta\) between \(\vec{m}\) and \(\vec{B}\), not the angle between the plane and \(\vec{B}\).
\( \textbf{Magnetic moment direction:} \) \(\vec{m}\) is perpendicular to the plane of the coil.
\( \textbf{Required angle:} \)
\[
\theta=90^\circ-30^\circ=60^\circ
\]
\( \textbf{Torque relation:} \)
\[
\tau=mB\sin\theta
\]
\( \textbf{Substitution:} \)
\[
\tau=(0.40)(0.50)\sin60^\circ
\]
\( \textbf{Use \(\sin60^\circ=\frac{\sqrt{3}}{2}\approx0.866\):} \)
\[
\tau=0.20(0.866)=0.173\,\text{N m}
\]
\( \textbf{Final answer:} \) The torque is \(0.173\,\text{N m}\).
315. Which row correctly distinguishes ammeter and voltmeter conversion of a galvanometer?
| Row | Ammeter conversion | Voltmeter conversion |
|---|
| P | Low shunt in parallel; final instrument in series | High resistance in series; final instrument in parallel |
| Q | High resistance in series; final instrument in parallel | Low shunt in parallel; final instrument in series |
| R | Low resistance in series; final instrument in parallel | High resistance in parallel; final instrument in series |
| S | High shunt in parallel; final instrument in series | Low resistance in series; final instrument in parallel |
ⓐ. Row Q
ⓑ. Row P
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: An ammeter measures current, so it must be connected in series with the circuit element. It should have very low resistance, so a low-resistance shunt is connected in parallel with the galvanometer to bypass most of the current. A voltmeter measures potential difference, so it must be connected in parallel with the circuit element. It should have very high resistance, so a high resistance is connected in series with the galvanometer. Row P correctly gives both conversions. The useful instrument-connection check is to remember both the added-resistance connection and how the final meter is connected in the circuit.
316. A current element produces magnetic field contribution \(dB\) at a point. If \(I\) is doubled, \(dl\) is unchanged, \(r\) is tripled, and \(\theta\) changes from \(90^\circ\) to \(30^\circ\), what is the new field contribution in terms of the original?
ⓐ. \(\frac{1}{6}dB\)
ⓑ. \(\frac{1}{3}dB\)
ⓒ. \(\frac{1}{9}dB\)
ⓓ. \(\frac{2}{9}dB\)
Correct Answer: \(\frac{1}{9}dB\)
Explanation: \( \textbf{Biot-Savart magnitude form:} \)
\[
dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}
\]
\( \textbf{Dependence:} \)
\[
dB\propto \frac{I\sin\theta}{r^2}
\]
\( \textbf{Current change:} \) \(I\) becomes \(2I\), so this contributes a factor \(2\).
\( \textbf{Angle change:} \) \(\sin90^\circ=1\), while \(\sin30^\circ=\frac{1}{2}\), so this contributes a factor \(\frac{1}{2}\).
\( \textbf{Distance change:} \) \(r\) becomes \(3r\), so the inverse-square factor contributes \(\frac{1}{9}\).
\( \textbf{Total factor:} \)
\[
2\times\frac{1}{2}\times\frac{1}{9}=\frac{1}{9}
\]
\( \textbf{Final answer:} \) The new field contribution is \(\frac{1}{9}dB\).
317. An Amperian loop is traversed anticlockwise as seen by an observer. By the right-hand convention, out of the page is positive. The loop encloses \(6\,\text{A}\) out of the page, \(4\,\text{A}\) into the page, and \(3\,\text{A}\) into the page. What is \(\oint\vec{B}\cdot d\vec{l}\)?
ⓐ. \(\mu_0(13\,\text{A})\)
ⓑ. \(-\mu_0(1\,\text{A})\)
ⓒ. \(\mu_0(7\,\text{A})\)
ⓓ. \(\mu_0(1\,\text{A})\)
Correct Answer: \(-\mu_0(1\,\text{A})\)
Explanation: \( \textbf{Ampere's law:} \)
\[
\oint\vec{B}\cdot d\vec{l}=\mu_0I_{\text{enc}}
\]
\( \textbf{Sign convention:} \) For anticlockwise traversal as seen by the observer, the positive normal is out of the page.
\( \textbf{Assign signs:} \) Out-of-page current is positive, and into-page current is negative.
\( \textbf{Algebraic current:} \)
\[
I_{\text{enc}}=6-4-3
\]
\( \textbf{Simplification:} \)
\[
I_{\text{enc}}=-1\,\text{A}
\]
\( \textbf{Line integral:} \)
\[
\oint\vec{B}\cdot d\vec{l}=\mu_0(-1\,\text{A})
\]
\( \textbf{Final answer:} \) The line integral is \(-\mu_0(1\,\text{A})\).
318. Two long parallel wires carry currents \(4I\) and \(I\) in opposite directions. The separation is \(d\). At which point on the line joining the wires can the magnetic field be zero?
ⓐ. Outside the two wires, on the side of the \(I\) wire
ⓑ. Between the two wires, closer to the \(4I\) wire
ⓒ. Between the two wires, closer to the \(I\) wire
ⓓ. Outside the two wires, on the side of the \(4I\) wire
Correct Answer: Outside the two wires, on the side of the \(I\) wire
Explanation: For opposite currents, the magnetic fields between the two wires are in the same direction, so they cannot cancel there. Cancellation is possible only outside the pair, where the two fields are in opposite directions. The field due to a long straight wire is \(B=\frac{\mu_0I}{2\pi r}\). To cancel the stronger \(4I\) current, the point must be farther from the \(4I\) wire and closer to the weaker \(I\) wire. That location is outside on the side of the \(I\) wire. The common mistake is to look only for the midpoint, but unequal currents shift the zero-field point toward the weaker current and outside for opposite current directions.
319. A long solenoid has internal magnetic field \(B\). Its number of turns is doubled and its length is also doubled, while current remains unchanged. What happens to the ideal internal field?
ⓐ. It becomes \(\frac{B}{2}\)
ⓑ. It becomes \(2B\)
ⓒ. It becomes \(4B\)
ⓓ. It remains \(B\)
Correct Answer: It remains \(B\)
Explanation: The magnetic field inside a long solenoid is
\[
B=\mu_0nI
\]
where
\[
n=\frac{N}{L}
\]
If both \(N\) and \(L\) are doubled, the turn density becomes:
\[
n'=\frac{2N}{2L}=\frac{N}{L}=n
\]
Since the current \(I\) is unchanged, the product \(nI\) remains unchanged. Therefore, the ideal internal magnetic field remains the same. The total number of turns alone is not enough to decide the solenoid field; turns per unit length is the relevant quantity.
320. A circular coil has \(N=25\) turns, radius \(R=0.20\,\text{m}\), and carries current \(I=0.50\,\text{A}\). A long straight wire carries current \(10\,\text{A}\). At what distance from the straight wire is its field equal to the field at the centre of the coil?
ⓐ. \(0.05\,\text{m}\)
ⓑ. \(0.10\,\text{m}\)
ⓒ. \(0.20\,\text{m}\)
ⓓ. \(0.40\,\text{m}\)
Correct Answer: \(0.05\,\text{m}\)
Explanation: \( \textbf{Centre field of coil:} \)
\[
B_c=\frac{\mu_0NI}{2R}
\]
\( \textbf{Field of long straight wire:} \)
\[
B_w=\frac{\mu_0I_w}{2\pi r}
\]
\( \textbf{Equal-field condition:} \)
\[
\frac{\mu_0NI}{2R}=\frac{\mu_0I_w}{2\pi r}
\]
\( \textbf{Cancel common factors:} \)
\[
\frac{NI}{R}=\frac{I_w}{\pi r}
\]
\( \textbf{Solve for \(r\):} \)
\[
r=\frac{I_wR}{\pi NI}
\]
\( \textbf{Substitution:} \)
\[
r=\frac{(10)(0.20)}{\pi(25)(0.50)}
\]
\( \textbf{Simplification:} \)
\[
r=\frac{2.0}{12.5\pi}=\frac{2.0}{39.25}\approx0.0509\,\text{m}
\]
\( \textbf{Rounded value:} \) \(r\approx0.05\,\text{m}\).
\( \textbf{Final answer:} \) The distance from the straight wire is \(0.05\,\text{m}\).
