Q1
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$d\vec{B} = \frac{\mu_0}{4\pi} \frac{i (d\vec{l} \times \vec{r})}{r^3}$
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.
$-\frac{56}{169} \times 10^{-9} \hat{k}\text{ T}$
Q2
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{12 K i}{5 r} (-\hat{k})$
Q3
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{K i}{2\sqrt{2} r} \otimes$
Q4
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{2 Ki}{5 r} \otimes$
Q5
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{\mu_0 i}{2\pi r} \otimes$
Q6
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{2 K i}{5 r} (-\hat{k})$
Q7
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{K i}{2 r} (-\hat{k})$
Q8
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$B_P = 0, B_Q = \frac{2 K i}{r}$
Q9
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$0$
Q10
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$2 K i \left[\frac{1}{r_1} + \frac{1}{r_2}\right] (\hat{k})$
Q11
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{K i_1}{r_1} (-\hat{k}) + \frac{K i_2}{r_2} (+\hat{k})$
Q12
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{K i}{r}$
Q13
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{K i}{r}$
Q14
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{K i}{2 r} \odot$
Q15
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{16 K i}{L} \otimes$
Q16
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{3\sqrt{3} K i}{l} \otimes$
Q17
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$14 K \odot$
Q18
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$0$
Q19
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{K i}{r}$
Q20
DPT
Biot Savart Law
MCQ
29 Jul 2026
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) $B_O = \frac{\mu_0 i}{2 R}$
Q21
DPT
Biot Savart Law
MCQ
29 Jul 2026
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) $B_O = \frac{K i}{R} \left(\frac{\pi}{3}\right)$
Q22
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$K i \frac{\pi}{3} \left[ \frac{1}{R_1} (+\hat{k}) + \frac{1}{R_2} (+\hat{k}) \right]$
Q23
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$0$
Q24
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$0$
Q25
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{\mu_0 i}{16 R} \odot$
Q26
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$0$
Q27
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{\mu_0 i}{8\pi R} \pi$
Q28
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$0$
Q29
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{2 K i}{R} \left( \frac{3\pi}{2} + 1 \right) \otimes$
Q30
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$0$
Q31
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{2 K i}{R} [-\hat{j} - \pi\hat{k}]$
Q32
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{3\sqrt{2} K i}{2 L}$
Q33
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$\frac{2 K i}{R} \pi$
Q34
DPT
Biot Savart Law
MCQ
29 Jul 2026
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.
$0$
Q35
DPT
EMI
MCQ
16 Aug 2026
Concept: The magnetic flux linked with a surface placed in a magnetic field is given by the dot product of the magnetic field vector and the area vector. The formula is $\phi = B A \cos\theta$, where $B$ is the magnetic field intensity, $A$ is the area, and $\theta$ is the angle between the normal to the plane of the coil (area vector) and the direction of the magnetic field.
A coil of area $A = 0.5\text{ m}^2$ is situated in a uniform magnetic field $B = 4.0\text{ wb/m}^2$ and area vector makes an angle of $60^\circ$ with respect to the magnetic field from the image given below. The value of the magnetic flux through the area $A$ would be equal to
A.
$2\text{ weber}$
B.
$1\text{ weber}$
C.
$3\text{ weber}$
D.
$\frac{3}{2}\text{ weber}$
Q36
DPT
EMI
MCQ
16 Aug 2026
Concept: The magnetic flux linked with a coil is given by $\phi = N B A \cos\theta$, where $N$ is the number of turns, $B$ is the magnetic field intensity, $A$ is the area, and $\theta$ is the angle between the magnetic field and the area vector. For the magnetic flux to be maximum, the cosine of the angle must be its maximum possible value, which is $\cos\theta = 1$. Thus, the maximum magnetic flux is $\phi_{\text{max}} = N B A$.
A coil of $N$ turns and area $A$ is rotated at the rate of $n$ rotations per second in a magnetic field of intensity $B$, the magnitude of the maximum magnetic flux will be from the image given below
A.
$NnAB$
B.
$NAB$
C.
$2\pi nNAB$
D.
$nAB$
Q37
DPT
EMI
MCQ
16 Aug 2026
Concept: The magnetic flux linked with an area placed in a uniform magnetic field is given by $\phi = B A \cos\theta$, where $B$ is the magnetic field intensity, $A$ is the area, and $\theta$ is the angle between the normal to the plane of the coil and the direction of the magnetic field. When the coil is placed perpendicular to the magnetic field, the angle $\theta = 0^\circ$.
A square coil of $10^{-2}\text{ m}^2$ area is placed perpendicular to a uniform magnetic field of intensity $10^3\text{ wb/m}^2$. The magnetic flux through the coil is
A.
$10^5\text{ weber}$
B.
$10\text{ weber}$
C.
$10^{-5}\text{ weber}$
D.
$100\text{ weber}$
Q38
DPT
EMI
MCQ
16 Aug 2026
Concept: The magnetic flux linked with a surface placed in a magnetic field is given by $\phi = B A \cos\theta$, where $B$ is the magnetic field intensity, $A$ is the area of the surface, and $\theta$ is the angle between the normal to the surface and the direction of the magnetic field.
Consider the following figure, a uniform magnetic field of $0.2\text{ T}$ is directed along the positive x-axis from the image given below. What is the magnetic flux through top surface of the figure
A.
$1.0\text{ m-wb}$
B.
Zero
C.
$0.8\text{ m-wb}$
D.
$-1.8\text{ m-wb}$
Q39
DPT
EMI
MCQ
16 Aug 2026
Concept: Faraday's second law of electromagnetic induction states that the induced emf in a circuit is equal to the negative rate of change of magnetic flux linked with the circuit. The formula is $e = -N \frac{\Delta\phi}{\Delta t} = -N A \frac{B_2 - B_1}{\Delta t} \cos\theta$, where $N$ is the number of turns, $A$ is the area, $B_1$ and $B_2$ are the initial and final magnetic fields, and $\Delta t$ is the time interval.
A coil of area $100\text{ cm}^2$ has $500$ turns. Magnetic field of $0.1\text{ weber/metre}^2$ is perpendicular to the coil from the image given below. The field is reduced to zero in $0.1\text{ sec}$. The induced emf in the coil is
A.
$1\text{ V}$
B.
$50\text{ V}$
C.
$5\text{ V}$
D.
Zero
Q40
DPT
EMI
MCQ
16 Aug 2026
Concept: The average induced emf in a coil rotated in a magnetic field is given by Faraday's law of electromagnetic induction: $e = -N \frac{\Delta\phi}{\Delta t} = -N \frac{B A (\cos\theta_2 - \cos\theta_1)}{\Delta t}$, where $N$ is the number of turns, $A$ is the area of the coil, $B$ is the magnetic induction field, and $\theta_1$ and $\theta_2$ are the initial and final angles between the area vector and the magnetic field.
A coil has $1000$ turns and $500\text{ cm}^2$ as it is area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{-5}\text{ wb/m}^2$. The coil is rotated through $180^\circ$ in $0.2\text{ sec}$. The average emf induced in the coil in $mV$ is from the image given below
A.
$15$
B.
$10$
C.
$5$
D.
$20$
Q41
DPT
EMI
MCQ
16 Aug 2026
Concept: According to Faraday's law of electromagnetic induction, the induced emf in a coil is given by the rate of change of magnetic flux linked with it. The formula is $e = -N \frac{\Delta B}{\Delta t} A \cos\theta$, where $N$ is the number of turns, $A$ is the area of each loop, $\frac{\Delta B}{\Delta t}$ is the rate of change of the magnetic field, and $\theta$ is the angle between the normal to the surface and the magnetic field.
A coil having $500$ square loops each of side $10\text{ cm}$ is placed normal to a magnetic field which increases at a rate of $1\text{ T/s}$ from the image given below. The induced emf in volt is
A.
$5$
B.
$0.1$
C.
$1$
D.
$0.5$
Q42
DPT
EMI
MCQ
16 Aug 2026
Concept: Faraday's law of electromagnetic induction states that the magnitude of the induced emf is equal to the rate of change of magnetic flux linked with the coil. The formula is $e = -N \frac{\Delta\phi}{\Delta t} = -N \frac{(B_2 - B_1) A \cos\theta}{\Delta t}$, where $N$ is the number of turns, $A$ is the area, $B_1$ and $B_2$ are the initial and final magnetic fields, $\theta$ is the angle, and $\Delta t$ is the time interval.
The magnetic field of $2 \times 10^{-2}\text{ Tesla}$ acts at right angle to a coil of area $100\text{ cm}^2$ with $50$ turns. The average emf induced in the coil is $0.1\text{ V}$ when it is removed from the field in time $t$. The value of $t$ is from the image given below
A.
$1\text{ s}$
B.
$0.1\text{ s}$
C.
$20\text{ s}$
D.
$0.01\text{ s}$
Q43
DPT
EMI
MCQ
16 Aug 2026
Concept: The charge induced in a circuit when the magnetic flux changes is given by $q = \frac{N}{R} \Delta\phi = \frac{N}{R} B A (\cos\theta_1 - \cos\theta_2)$, where $N$ is the number of turns, $R$ is the total resistance, $B$ is the magnetic induction, $A$ is the area, and $\theta_1$ and $\theta_2$ are the initial and final angles of the coil with respect to the magnetic field.
A circular coil of $500$ turns of a wire has an enclosed area of $0.1\text{ m}^2$ per turn. It is kept perpendicular to a magnetic field of induction $0.2\text{ T}$ and rotated by $180^\circ$ about a diameter perpendicular to the field in $0.1\text{ sec}$ from the image given below. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of $50\text{ ohms}$
A.
$0.2\text{ C}$
B.
$0.4\text{ C}$
C.
$2\text{ C}$
D.
$4\text{ C}$
Q44
DPT
EMI
MCQ
16 Aug 2026
Concept: Faraday's law of electromagnetic induction states that the induced emf is equal to the negative rate of change of magnetic flux, $e = - \frac{d\phi}{dt}$. The induced current is given by Ohm's law as $i = \frac{e}{R} = - \frac{1}{R} \frac{d\phi}{dt}$, where $R$ is the resistance of the circuit.
Flux $\phi$ (in weber) in a closed circuit of resistance $10\text{ ohm}$ varies with time $t$ (in sec) according to the equation $\phi = 6t^2 - 5t + 1$ from the image given below. What is the magnitude of the induced current at $t = 0.25\text{ s}$
A.
$1.2\text{ A}$
B.
$0.8\text{ A}$
C.
$0.6\text{ A}$
D.
$0.2\text{ A}$
Q45
DPT
EMI
MCQ
16 Aug 2026
Concept: As the magnet moves towards the coil, the magnetic flux increases nonlinearly and an emf is induced. When the magnet passes through and moves away to the other side of the coil, the flux decreases and there is a change in polarity of the induced emf.
The variation of induced emf $E$ with time $t$ in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as from the image given below
A.
B.
C.
D.
Q46
DPT
EMI
MCQ
16 Aug 2026
Concept: The induced charge in a loop is given by $q = \frac{\vert{}e\vert{}}{R} \Delta t = \frac{\Delta\phi}{R} = \frac{B(A_2 - A_1)}{R}$, where $A_1$ is the initial area of the square loop and $A_2$ is the final area of the circular loop formed.
A square loop of side $a$ and resistance $R$ is placed in a transverse uniform magnetic field $B$. If it suddenly changes into circular form in time $t$ from the image given below, then magnitude of induced charge will be
A.
$\frac{Ba^2}{R}(4\pi - 1)$
B.
$\frac{Ba^2}{R}\left(1 - \frac{1}{4\pi}\right)$
C.
$\frac{Ba^2}{R}\left(\frac{1}{4\pi} - 1\right)$
D.
$\frac{Ba^2}{R}\left(\frac{4}{\pi} - 1\right)$
Q47
DPT
EMI
MCQ
16 Aug 2026
Concept: According to Faraday's law of electromagnetic induction, an emf is induced in a circuit whenever there is a change in magnetic flux linked with it, which typically results from relative motion between the source and the coil. The relative speed between the coil and the magnet determines the rate of change of flux: $e = -N \frac{d\phi}{dt}$. If the relative speed is zero, the magnetic flux remains constant with time, resulting in zero induced emf.
A circular coil and a bar magnet placed near by are made to move in the same direction from the image given below. The coil covers a distance of $1\text{ m}$ in $0.5\text{ sec}$ and the magnet a distance of $2\text{ m}$ in $1\text{ sec}$. The induced emf produced in the coil
A.
$1\text{ V}$
B.
$0.5\text{ V}$
C.
Zero
D.
Cannot be determined from the given information
Q48
DPT
EMI
MCQ
16 Aug 2026
Concept: The electrical power dissipated in a circuit is given by $P = \frac{e^2}{R}$, where $e$ is the induced emf and $R$ is the resistance of the wire. The induced emf is given by $e = -N A \frac{dB}{dt}$, and the resistance of a wire of length $l$ and radius $r$ is $R \propto \frac{l}{r^2}$. Since length $l$ is proportional to the number of turns $N$ for a constant area of each turn, the power scales as $P \propto Nr^2$.
A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would from the image given below
A.
Halved
B.
The same
C.
Doubled
D.
Quadrupled
Q49
DPT
EMI
MCQ
16 Aug 2026
Concept: The induced emf in a loop due to a changing area in a uniform magnetic field is given by Faraday's law: $e = - \frac{d\phi}{dt} = -B \frac{dA}{dt}$. Since the area of a circular loop is $A = \pi r^2$, the rate of change of area is $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$, which gives the induced emf formula $e = -B (2\pi r) \frac{dr}{dt}$.
A conducting circular loop is placed in a uniform magnetic field $B = 40\text{ mT}$ with its plane perpendicular to the field. If the radius of the loop starts shrinking at a constant rate $0.2\text{ mm/s}$, then the induced emf in the loop at an instant when its radius is $1.0\text{ cm}$ is
A.
$0.1\pi\text{ }\mu\text{V}$
B.
$0.2\pi\text{ }\mu\text{V}$
C.
$1.0\pi\text{ }\mu\text{V}$
D.
$0.16\pi\text{ }\mu\text{V}$
Q50
DPT
EMI
MCQ
16 Aug 2026
Concept: The magnetic field at the centre of a solenoid is given by $B = \mu_0 n i = \mu_0 \frac{N i}{l}$, where $\mu_0$ is the permeability of free space, $N$ is the number of turns, $l$ is the length, and $i$ is the current. The induced emf in a coaxial secondary coil of $N^\prime$ turns and area $A$ is given by Faraday's law: $e = -N^\prime A \frac{dB}{dt} = -N^\prime A \frac{-B - B}{\Delta t} = \frac{2N^\prime B A}{\Delta t}$.
A solenoid has $2000$ turns wound over a length of $0.314\text{ m}$. Around its central section a coil of $100$ turns and area of cross-section $1 \times 10^{-3}\text{ m}^2$ is wound. If an initial current of $2\text{ A}$ in the solenoid is reversed in $0.25\text{ sec}$, the emf induced in the coil is equal to
A.
$6 \times 10^{-4}\text{ V}$
B.
$12.8\text{ mV}$
C.
$6 \times 10^{-2}\text{ V}$
D.
$12.8\text{ V}$