Question 1: Palette 1/34

DPT MCQ

Concept: According to the Biot-Savart Law, the magnetic field $d\vec{B}$ due to a current element $i d\vec{l}$ at a position vector $\vec{r}$ is given by:
$d\vec{B} = \frac{\mu_0}{4\pi} \frac{i (d\vec{l} \times \vec{r})}{r^3}$

Find the magnetic field due to a small element $d\vec{l} = (3\hat{i} + 4\hat{j})\text{ mm}$ having current $10\text{ A}$ through it at a point whose position vector is $(5\hat{i} - 12\hat{j})\text{ m}$ from this element.

A

$-\frac{56}{2197} \times 10^{-9} \hat{k}\text{ T}$

B

$\frac{56}{2197} \times 10^{-9} \hat{k}\text{ T}$

C

$-\frac{28}{2197} \times 10^{-9} \hat{k}\text{ T}$

D

$-\frac{56}{169} \times 10^{-9} \hat{k}\text{ T}$

Question 2: Palette 2/34

DPT MCQ

Concept: The magnetic field due to a finite straight current-carrying wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4\pi r} (\sin\theta_1 + \sin\theta_2) = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2)$, directed into the page ($-\hat{k}$) according to the right-hand thumb rule.

From the image given below, find the magnetic field $B_P$ at point P due to a straight wire carrying current $i$ at a perpendicular distance $r$, where the angles subtended by the upper and lower ends of the wire at point P are $37^\circ$ and $53^\circ$ respectively.

A

$\frac{7 K i}{5 r} (-\hat{k})$

B

$\frac{1.4 K i}{r} (\hat{k})$

C

$\frac{K i}{5 r} (-\hat{k})$

D

$\frac{12 K i}{5 r} (-\hat{k})$

Question 3: Palette 3/34

DPT MCQ

Concept: The magnetic field due to a finite straight current-carrying wire at a perpendicular distance $r$ is given by the Biot-Savart formula $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) \otimes$, where $K = \frac{\mu_0}{4\pi} = 10^{-7} \text{ T}\cdot\text{m/A}$ and $\otimes$ represents the direction perpendicularly into the page.

From the image gven below, find the magnetic field $B_P$ at point P due to a straight wire carrying current $i$ at a perpendicular distance $r$, where the angles subtended by both ends of the wire at point P are $45^\circ$ and $45^\circ$ respectively.

A

$\frac{K i}{\sqrt{2} r} \otimes$

B

$\frac{2 K i}{r} \otimes$

C

$\frac{\sqrt{2} K i}{r} \otimes$

D

$\frac{K i}{2\sqrt{2} r} \otimes$

Question 4: Palette 4/34

DPT MCQ

Concept: The magnetic field due to a straight current-carrying wire at a perpendicular distance $r$ is given by $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) \otimes$, where $K = \frac{\mu_0}{4\pi} = 10^{-7} \text{ T}\cdot\text{m/A}$ and $\otimes$ indicates the direction perpendicularly into the page.

From the image gven below, find the magnetic field $B_P$ at point P due to a current-carrying wire of current $i$ where one end subtends an angle $\theta_1 = 53^\circ$ and the other end is directly opposite to point P ($\theta_2 = 0^\circ$) at a perpendicular distance $r$.

A

$\frac{3 Ki}{5 r} \otimes$

B

$\frac{4 Ki}{5 r} \otimes$

C

$\frac{Ki}{r} \otimes$

D

$\frac{2 Ki}{5 r} \otimes$

Question 5: Palette 5/34

DPT MCQ

Concept: The magnetic field due to a straight current-carrying wire at a perpendicular distance $r$ is given by $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) \otimes$, where $K = \frac{\mu_0}{4\pi} = 10^{-7} \text{ T}\cdot\text{m/A}$ and $\otimes$ indicates direction perpendicularly into the page.

From the image gven below, find the magnetic field $B_P$ at point P due to a semi-infinite straight wire carrying current $i$ at a perpendicular distance $r$, where one end extends to infinity ($\theta_1 = 90^\circ$) and the other end is directly opposite to point P ($\theta_2 = 0^\circ$).

A

$\frac{2 K i}{r} \otimes$

B

$\frac{K i}{2 r} \otimes$

C

$\frac{K i}{r} \otimes$

D

$\frac{\mu_0 i}{2\pi r} \otimes$

Question 6: Palette 6/34

DPT MCQ

Concept: The magnetic field due to a straight wire carrying current $i$ at a perpendicular distance $r$ is given by $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) (-\hat{k})$, where $K = \frac{\mu_0}{4\pi}$ and $(-\hat{k})$ denotes direction perpendicularly into the page.

From the image gven below, find the magnetic field $B_P$ at point P due to a semi-infinite straight wire carrying current $i$ at a perpendicular distance $r$, where one end extends to infinity ($\theta_1 = 90^\circ$) and the other end subtends an angle $\theta_2 = 37^\circ$.

A

$\frac{3 K i}{5 r} (-\hat{k})$

B

$\frac{8 K i}{5 r} (-\hat{k})$

C

$\frac{4 K i}{5 r} (-\hat{k})$

D

$\frac{2 K i}{5 r} (-\hat{k})$

Question 7: Palette 7/34

DPT MCQ

Concept: The magnetic field due to an infinitely long straight current-carrying wire at a perpendicular distance $r$ is given by $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) (-\hat{k}) = \frac{2 K i}{r} (-\hat{k}) = \frac{\mu_0 i}{2\pi r} (-\hat{k})$, where $K = \frac{\mu_0}{4\pi}$ and $(-\hat{k})$ indicates direction perpendicularly into the page.

From the image gven below, find the magnetic field $B_P$ at point P due to an infinitely long straight wire carrying current $i$ at a perpendicular distance $r$, where both ends extend to infinity ($\theta_1 = 90^\circ$ and $\theta_2 = 90^\circ$).

A

$\frac{K i}{r} (-\hat{k})$

B

$\frac{2 K i}{r} (-\hat{k})$

C

$\frac{4 K i}{r} (-\hat{k})$

D

$\frac{K i}{2 r} (-\hat{k})$

Question 8: Palette 8/34

DPT MCQ

Concept: According to the Biot-Savart Law, the magnetic field is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{i (d\vec{l} \times \vec{r})}{r^3}$. When a point lies along the axis of the current-carrying wire, the angle $\theta$ between the current element $d\vec{l}$ and the position vector $\vec{r}$ is either $0^\circ$ or $180^\circ$, making $d\vec{l} \times \vec{r} = 0$ and thus the magnetic field $B = 0$.

From the image gven below, find the magnetic field at points P and Q which lie on the axis (extended line) of a straight wire carrying current $i$.

A

$B_P = 0, B_Q = 0$

B

$B_P = \frac{K i}{r}, B_Q = \frac{K i}{r}$

C

$B_P = \frac{2 K i}{r}, B_Q = 0$

D

$B_P = 0, B_Q = \frac{2 K i}{r}$

Question 9: Palette 9/34

DPT MCQ

Concept: The magnetic field due to a straight wire carrying current $i$ at a perpendicular distance $a$ is given by $\vec{B} = \frac{K i}{a} (\sin\theta_1 + \sin\theta_2) \hat{n}$, where $K = \frac{\mu_0}{4\pi}$. The magnetic field contribution at any point lying along the axis of a current-carrying wire is zero.

From the image gven below, find the magnetic field vector $\vec{B}_P$ at point P $(0,0)$ due to the wire arrangement consisting of a horizontal current-carrying wire lying on the x-axis and a vertical semi-infinite wire carrying current $i$ extending upwards to infinity from $(-a, 0)$.

A

$\frac{2 K i}{a} (-\hat{k})$

B

$\frac{K i}{a} (-\hat{k})$

C

$\frac{K i}{2 a} (-\hat{k})$

D

$0$

Question 10: Palette 10/34

DPT MCQ

Concept: The magnetic field due to an infinitely long straight wire carrying current $i$ at a perpendicular distance $r$ is given by $B = \frac{2 K i}{r}$, where $K = \frac{\mu_0}{4\pi}$. By the right-hand thumb rule, parallel wires carrying currents in opposite directions produce magnetic fields in the same direction at a point located between them.

From the image gven below, find the total magnetic field vector $\vec{B}_A$ at point P situated between two infinitely long parallel straight wires carrying currents $i$ in opposite directions at perpendicular distances $r_1$ and $r_2$ respectively from point P.

A

$2 K i \left[\frac{1}{r_1} - \frac{1}{r_2}\right] (-\hat{k})$

B

$2 K i \left[\frac{1}{r_1} + \frac{1}{r_2}\right] (-\hat{k})$

C

$K i \left[\frac{1}{r_1} + \frac{1}{r_2}\right] (-\hat{k})$

D

$2 K i \left[\frac{1}{r_1} + \frac{1}{r_2}\right] (\hat{k})$

Question 11: Palette 11/34

DPT MCQ

Concept: The magnetic field due to an infinitely long straight wire carrying current $i$ at a perpendicular distance $r$ is given by $B = \frac{2 K i}{r}$, where $K = \frac{\mu_0}{4\pi}$. The direction of the magnetic field vector is determined by the right-hand thumb rule.

From the image gven below, find the net magnetic field vector $\vec{B}_P$ at point P situated between two infinitely long parallel straight wires carrying currents $i_1$ and $i_2$ both in the upward direction at perpendicular distances $r_1$ and $r_2$ respectively from point P.

A

$\frac{2 K i_1}{r_1} (-\hat{k}) + \frac{2 K i_2}{r_2} (-\hat{k})$

B

$\frac{2 K i_1}{r_1} (+\hat{k}) + \frac{2 K i_2}{r_2} (-\hat{k})$

C

$\frac{2 K i_1}{r_1} (-\hat{k}) + \frac{2 K i_2}{r_2} (+\hat{k})$

D

$\frac{K i_1}{r_1} (-\hat{k}) + \frac{K i_2}{r_2} (+\hat{k})$

Question 12: Palette 12/34

DPT MCQ

Concept: The magnetic field due to an infinitely long straight wire carrying current $i$ at a perpendicular distance $r$ is given by $B = \frac{2 K i}{r}$, where $K = \frac{\mu_0}{4\pi}$. By the right-hand thumb rule, equal currents in the same direction produce equal and opposite magnetic fields at the midpoint between them, resulting in vector cancellation.

From the image gven below, find the net magnetic field $B_P$ at point P situated midway between two long parallel straight wires, each carrying equal current $i$ in the same direction, separated by a distance $2r$ (such that each wire is at a distance $r$ from point P).

A

$\frac{4 K i}{r}$

B

$\frac{2 K i}{r}$

C

$0$

D

$\frac{K i}{r}$

Question 13: Palette 13/34

DPT MCQ

Concept: The magnetic field due to an infinitely long straight wire carrying current $i$ at a perpendicular distance $r$ is given by $B = \frac{2 K i}{r}$, where $K = \frac{\mu_0}{4\pi}$. When two parallel wires carry currents in opposite directions, their magnetic field contributions at any point situated between them act in the same direction, adding together.

From the image gven below, find the total magnitude of the magnetic field $B_P$ at point P situated midway between two infinitely long parallel straight wires carrying equal current $i$ in opposite directions, where each wire is at a perpendicular distance $r$ from point P.

A

$\frac{2 K i}{r}$

B

$0$

C

$\frac{4 K i}{r}$

D

$\frac{K i}{r}$

Question 14: Palette 14/34

DPT MCQ

Concept: The magnetic field due to a semi-infinite straight current-carrying wire at a perpendicular distance $r$, when the point is level with one end ($\theta_1 = 90^\circ, \theta_2 = 0^\circ$), is given by $B = \frac{K i}{r}$, where $K = \frac{\mu_0}{4\pi}$. By the right-hand thumb rule, semi-infinite wires with opposite currents produce magnetic fields in opposite directions at a point situated centrally below them.

From the image gven below, find the net magnetic field $B_P$ at point P situated below two semi-infinite parallel straight wires, each extending upwards to infinity, carrying current $i$ in opposite directions at equal horizontal distances $r$ from point P, where point P lies on the line passing through their lower ends.

A

$\frac{2 K i}{r} \otimes$

B

$0$

C

$\frac{K i}{r} \otimes$

D

$\frac{K i}{2 r} \odot$

Question 15: Palette 15/34

DPT MCQ

Concept: The magnetic field due to a single straight segment carrying current $i$ at a perpendicular distance $r = \frac{L}{2}$ with subtended angles $\theta_1 = 45^\circ$ and $\theta_2 = 45^\circ$ is $B_1 = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2)$. For a square loop consisting of 4 identical sides, the net magnetic field at the center is $B_P = 4 \times \frac{K i}{L/2} (\sin 45^\circ + \sin 45^\circ)$, directed perpendicularly into the page.

From the image gven below, find the total magnetic field $B_P$ at the center P of a square loop of side length $L$ carrying current $i$.

A

$\frac{4\sqrt{2} K i}{L} \otimes$

B

$\frac{8\sqrt{2} K i}{L} \otimes$

C

$\frac{2\sqrt{2} K i}{L} \otimes$

D

$\frac{16 K i}{L} \otimes$

Question 16: Palette 16/34

DPT MCQ

Concept: The magnetic field at the centroid P of an equilateral triangular loop is the sum of the magnetic fields due to its three equal sides. For each side, the perpendicular distance from the center is $r = \frac{l}{2\sqrt{3}}$, and the angles subtended by the ends are $\theta_1 = 60^\circ$ and $\theta_2 = 60^\circ$. The net magnetic field is given by $B_P = 3 \times \frac{K i}{r} (\sin 60^\circ + \sin 60^\circ)$.

From the image gven below, find the total magnetic field $B_P$ at the center P of an equilateral triangular loop of side length $l$ carrying current $i$.

A

$\frac{18 K i}{l} \otimes$

B

$\frac{9 K i}{l} \otimes$

C

$\frac{6\sqrt{3} K i}{l} \otimes$

D

$\frac{3\sqrt{3} K i}{l} \otimes$

Question 17: Palette 17/34

DPT MCQ

Concept: The magnetic field due to an infinitely long straight wire carrying current $i$ at a perpendicular distance $r$ is given by $B = \frac{2 K i}{r}$, where $K = \frac{\mu_0}{4\pi}$. The net magnetic field at any point is the vector sum of individual fields determined using the right-hand thumb rule.

From the image gven below, find the net magnetic field $B_P$ at point P with coordinates $(3, 4)$ due to two infinitely long perpendicular wires lying along the coordinate axes, carrying currents $12\text{ A}$ along the y-axis (upward) and $4\text{ A}$ along the x-axis (leftward).

A

$6 K \odot$

B

$10 K \odot$

C

$6 K \otimes$

D

$14 K \odot$

Question 18: Palette 18/34

DPT MCQ

Concept: The magnetic field due to a straight current-carrying wire segment at a perpendicular distance $r$ is given by $B = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2)$, where $K = \frac{\mu_0}{4\pi}$. If a point lies along the axis of a straight current-carrying wire, the magnetic field contribution from that wire segment is zero. The net magnetic field is the vector sum of contributions from all wire segments.

From the image gven below, find the total magnetic field vector at point O due to the given current-carrying wire configuration carrying current $i$.

A

$\frac{2 K i}{r} (-\hat{k})$

B

$\frac{K i}{r} (-\hat{k})$

C

$\frac{4 K i}{r} (-\hat{k})$

D

$0$

Question 19: Palette 19/34

DPT MCQ

Concept: When a point A lies outside the projection of the wire PQ, one end subtends an angle $\theta_1$ and the other subtends an angle $\theta_2$ on the same side of the perpendicular. In this geometry, $\theta_2$ is taken as negative ($-\theta_2$), so the magnetic field is given by $B_A = \frac{K i}{r} (\sin\theta_1 - \sin\theta_2)$, where $K = \frac{\mu_0}{4\pi}$.

From the image gven below, find the magnetic field $B_A$ at point A due to a straight wire carrying current $i$ where both ends P and Q lie on the same side of the perpendicular drawn from point A to the line containing the wire (extended line), subtending angles $53^\circ$ and $37^\circ$ with the perpendicular line of length $r$.

A

$\frac{K i}{5 r}$

B

$\frac{7 K i}{5 r}$

C

$\frac{2 K i}{5 r}$

D

$\frac{K i}{r}$

Question 20: Palette 20/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc carrying current $i$ subtending an angle $\theta$ at its center is given by $B_O = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$ and $\theta$ is in radians. For a complete circular loop ($\theta = 2\pi$), $B_O = \frac{\mu_0 i}{2 R}$, and for a semi-circular arc ($\theta = \pi$), $B_O = \frac{K i}{R} \pi = \frac{\mu_0 i}{4 R}$.

From the image gven below, find the magnetic field $B_O$ at the center O for the given current configurations: (a) a complete circular loop of radius $R$ carrying current $i$, and (b) a semi-circular arc of radius $R$ carrying current $i$.

A

(a) $B_O = \frac{\mu_0 i}{2 R}$, (b) $B_O = \frac{K i}{R} \pi$

B

(a) $B_O = \frac{\mu_0 i}{4 R}$, (b) $B_O = \frac{2 K i}{R} \pi$

C

(a) $B_O = \frac{2 K i}{R}$, (b) $B_O = \frac{K i}{2 R} \pi$

D

(a) $B_O = \frac{K i}{R} \pi$, (b) $B_O = \frac{\mu_0 i}{2 R}$

Question 21: Palette 21/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ carrying current $i$ that subtends an angle $\theta$ (in radians) at the center is given by $B_O = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$.

From the image gven below, find the magnetic field $B_O$ at the center O for the given current-carrying circular arcs: (a) a quarter-circular arc subtending an angle of $90^\circ$ at center O, and (b) a circular arc subtending an angle of $60^\circ$ at center O, both having radius $R$ and carrying current $i$.

A

(a) $B_O = \frac{K i}{R} \left(\frac{\pi}{2}\right)$, (b) $B_O = \frac{K i}{R} \left(\frac{\pi}{3}\right)$

B

(a) $B_O = \frac{K i}{R} \pi$, (b) $B_O = \frac{K i}{R} \left(\frac{\pi}{2}\right)$

C

(a) $B_O = \frac{2 K i}{R} \left(\frac{\pi}{2}\right)$, (b) $B_O = \frac{K i}{R} \left(\frac{\pi}{6}\right)$

D

(a) $B_O = \frac{K i}{R} \left(\frac{\pi}{4}\right)$, (b) $B_O = \frac{K i}{R} \left(\frac{\pi}{3}\right)$

Question 22: Palette 22/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ subtending angle $\theta$ in radians carrying current $i$ is $B = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. Radial straight wire segments whose lines of action pass through the center O contribute zero magnetic field at O.

From the image gven below, find the net magnetic field vector $\vec{B}_O$ at center O due to the closed loop consisting of two concentric circular arcs of radii $R_1$ and $R_2$ subtending an angle of $60^\circ$ at O, and two radial straight wire segments carrying current $i$.

A

$K i \frac{\pi}{3} \left[ \frac{1}{R_1} (-\hat{k}) + \frac{1}{R_2} (+\hat{k}) \right]$

B

$K i \frac{\pi}{6} \left[ \frac{1}{R_1} (-\hat{k}) + \frac{1}{R_2} (+\hat{k}) \right]$

C

$K i \frac{\pi}{3} \left[ \frac{1}{R_1} (+\hat{k}) + \frac{1}{R_2} (-\hat{k}) \right]$

D

$K i \frac{\pi}{3} \left[ \frac{1}{R_1} (+\hat{k}) + \frac{1}{R_2} (+\hat{k}) \right]$

Question 23: Palette 23/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ carrying current $i$ subtending an angle $\theta$ (in radians) is given by $B = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. Straight wire segments whose extensions pass directly through the point O contribute zero magnetic field at O.

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at point O due to the wire configuration consisting of a semi-circular arc of radius $R$ and two horizontal straight wire segments extending along the line passing through point O, carrying a current $i$.

A

$\frac{K i}{R} \pi \otimes$

B

$\frac{2 K i}{R} \pi \otimes$

C

$\frac{K i}{2 R} \pi \odot$

D

$0$

Question 24: Palette 24/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ carrying current $i$ subtending an angle $\theta$ (in radians) is given by $B = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. Straight wire segments whose lines of action pass directly through the point O contribute zero magnetic field at O.

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at origin O due to the current-carrying wire configuration consisting of a horizontal straight wire segment along the x-axis, a quarter-circular arc of radius $R$, and a vertical straight wire segment along the y-axis extending to infinity, carrying current $i$.

A

$\frac{K i}{2 R} \pi \odot$

B

$\frac{K i}{R} \frac{\pi}{2} \odot$

C

$\frac{2 K i}{R} \pi \otimes$

D

$0$

Question 25: Palette 25/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ carrying current $i$ subtending an angle $\theta$ in radians is given by $B = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. The net magnetic field at the common center O is the vector sum of the magnetic fields produced by each arc.

From the image gven below, find the net magnetic field $B_O$ at the center O due to two concentric semi-circular arcs of radii $R$ and $2R$ carrying current $i$ in opposite directions as shown.

A

$\frac{\mu_0 i}{8 R} \odot$

B

$\frac{\mu_0 i}{4 R} \odot$

C

$\frac{\mu_0 i}{8 R} \otimes$

D

$\frac{\mu_0 i}{16 R} \odot$

Question 26: Palette 26/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ carrying current $i$ subtending an angle $\theta$ (in radians) is given by $B = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. Straight wire segments whose lines of action pass directly through point O contribute zero magnetic field at O.

From the image gven below, find the total magnetic field $B_O$ at point O due to a wire configuration consisting of a circular arc of radius $R$ subtending an angle of $60^\circ$ carrying current $i$, along with two radial straight wire segments extending from point O.

A

$\frac{K i}{R} \frac{\pi}{3} \otimes$

B

$\frac{K i}{R} \frac{\pi}{6} \otimes$

C

$\frac{2 K i}{R} \frac{\pi}{3} \otimes$

D

$0$

Question 27: Palette 27/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ carrying current $i$ subtending an angle $\theta$ (in radians) is given by $B_O = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. Substituting $K$ into the formula gives $B_O = \frac{\mu_0 i}{4\pi R} \theta$.

From the image gven below, find the total magnetic field $B_O$ at the center O due to a major circular arc wire segment of radius $R$ carrying current $i$ that subtends an angle $\theta = \frac{3\pi}{2}$ radians at center O.

A

$\frac{\mu_0 i}{4\pi R} \frac{3\pi}{2}$

B

$\frac{\mu_0 i}{2\pi R} \frac{3\pi}{2}$

C

$\frac{\mu_0 i}{4\pi R} \frac{\pi}{2}$

D

$\frac{\mu_0 i}{8\pi R} \pi$

Question 28: Palette 28/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ carrying current $i$ subtending an angle $\theta$ (in radians) is given by $B = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. Straight wire segments whose lines of action pass directly through point O contribute zero magnetic field at O.

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at point O due to the current-carrying wire configuration consisting of a circular arc of radius $R$ subtending an angle of $\frac{3\pi}{2}$ radians at center O, and two semi-infinite straight wires extending along lines passing through point O, carrying current $i$.

A

$\frac{K i}{R} \frac{3\pi}{2} \odot$

B

$\frac{K i}{R} \pi \otimes$

C

$\frac{2 K i}{R} \frac{3\pi}{2} \otimes$

D

$0$

Question 29: Palette 29/34

DPT MCQ

Concept: The magnetic field due to a circular arc at its center is $B_{\text{arc}} = \frac{K i}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. The magnetic field due to a straight wire at perpendicular distance $r$ with subtended angles $\theta_1$ and $\theta_2$ is $B_{\text{wire}} = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2)$. The net magnetic field is the vector sum of all individual contributions.

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at point O due to the current-carrying wire configuration consisting of a circular arc of radius $R$ subtending an angle of $\frac{3\pi}{2}$ radians at center O, a semi-infinite straight wire whose line of action passes through O, and a semi-infinite straight wire running parallel to the horizontal at a distance $R$ from O with $\theta_1 = 90^\circ$ and $\theta_2 = 0^\circ$, carrying current $i$.

A

$\frac{K i}{R} \left( \frac{3\pi}{2} + 1 \right) \odot$

B

$\frac{K i}{R} \left( \frac{3\pi}{2} - 1 \right) \otimes$

C

$\frac{K i}{R} \left( \frac{\pi}{2} + 1 \right) \odot$

D

$\frac{2 K i}{R} \left( \frac{3\pi}{2} + 1 \right) \otimes$

Question 30: Palette 30/34

DPT MCQ

Concept: The magnetic field at the center O of a square loop of side length $L$ is equal to 3 times the magnetic field due to a single side carrying current $i$, as the field from the 3 sides adds up in the same direction. The distance from the center O to each side is $r = \frac{L}{2}$, and the angles subtended by each side at the center are $\theta_1 = 45^\circ$ and $\theta_2 = 45^\circ$. The magnetic field due to one side is $B_1 = \frac{K i}{r} (\sin 45^\circ + \sin 45^\circ)$.

From the image gven below, find the total magnetic field $B_O$ at the center O of a square loop of side length $L$ carrying current $i$ where current enters and leaves through long diagonal leads.

A

$\frac{3\sqrt{2} K i}{L/2} \otimes$

B

$\frac{\sqrt{2} K i}{L/2} \otimes$

C

$\frac{2\sqrt{2} K i}{L/2} \otimes$

D

$0$

Question 31: Palette 31/34

DPT MCQ

Concept: The magnetic field vector due to a semi-infinite straight wire at perpendicular distance $R$ is given by $B = \frac{K i}{R}$, and due to a semi-circular arc of radius $R$ is $B = \frac{K i}{R} \pi$, where $K = \frac{\mu_0}{4\pi}$. The direction of each field contribution is determined by the right-hand rule, and the net magnetic field vector is the vector sum $\vec{B}_O = \vec{B}_1 + \vec{B}_2 + \vec{B}_3$.

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at point O due to the current-carrying wire configuration consisting of two semi-infinite straight wires and a semi-circular arc of radius $R$ in the xy-plane carrying current $i$.

A

$\frac{K i}{R} [-2\hat{j} - \pi\hat{k}]$

B

$\frac{K i}{R} [2\hat{j} + \pi\hat{k}]$

C

$\frac{K i}{R} [-\hat{j} - \pi\hat{k}]$

D

$\frac{2 K i}{R} [-\hat{j} - \pi\hat{k}]$

Question 32: Palette 32/34

DPT MCQ

Concept: For a uniform conductor forming a closed symmetrical loop, when current $i$ enters at one point and exits at another, the current divides into two parallel branches inversely proportional to their resistances. The magnetic field at the center of a regular polygon due to one side carrying current $I$ is given by $B = \frac{K I}{L/2} (\sin 45^\circ + \sin 45^\circ) = \frac{2\sqrt{2} K I}{L}$, where $K = \frac{\mu_0}{4\pi}$.

From the image gven below, find the total magnetic field $B_O$ at the center O of a square loop of side length $L$ and uniform resistance per unit length, when a total current $i$ enters at one corner and leaves at an adjacent corner.

A

$\frac{2\sqrt{2} K i}{L}$

B

$0$

C

$\frac{\sqrt{2} K i}{2 L}$

D

$\frac{3\sqrt{2} K i}{2 L}$

Question 33: Palette 33/34

DPT MCQ

Concept: The magnetic field at the center of a circular arc of radius $R$ subtending angle $\theta$ carrying current $I$ is given by $B = \frac{K I}{R} \theta$, where $K = \frac{\mu_0}{4\pi}$. When two identical semi-circular arcs carry equal currents in opposite directions around the center, their magnetic fields at the center are equal in magnitude and opposite in direction, cancelling each other out. Straight lead wires whose lines of action pass through the center O contribute zero magnetic field at O.

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at the center O of a circular loop of radius $R$ where current $i$ enters at one point, splits equally into two semi-circular paths carrying current $i/2$ each, and exits at the diametrically opposite point.

A

$0$

B

$\frac{K i}{R} \pi \otimes$

C

$\frac{K i}{2 R} \pi \odot$

D

$\frac{2 K i}{R} \pi$

Question 34: Palette 34/34

DPT MCQ

Concept: The magnetic field at the center of a circular loop of radius $R$ carrying current $i$ is $B_{\text{loop}} = \frac{2 K i}{R} = \frac{\mu_0 i}{2 R}$, where $K = \frac{\mu_0}{4\pi}$. The magnetic field due to an infinitely long straight wire carrying current $i$ at a perpendicular distance $R$ is $B_{\text{wire}} = \frac{\mu_0 i}{2\pi R}$. The net magnetic field is the vector sum $\vec{B}_O = \vec{B}_{\text{loop}} + \vec{B}_{\text{wire}}$.

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at the center O due to the current-carrying wire system consisting of a complete circular loop of radius $R$ carrying current $i$ and an infinitely long straight wire tangent to the bottom of the loop carrying current $i$ in the opposite direction.

A

$\left( \frac{2 K i}{R} - \frac{\mu_0 i}{2\pi R} \right) \hat{k}$

B

$\left( \frac{K i}{R} + \frac{\mu_0 i}{2\pi R} \right) \hat{k}$

C

$\left( \frac{2 K i}{R} + \frac{\mu_0 i}{\pi R} \right) \hat{k}$

D

$0$