1
KCET 2026
MCQ (Single Correct Answer)
+1
-0
Two identical circular current loops carrying equal currents are placed with their axes inclined at $45^\circ$ to each other as shown in the figure. The resultant magnetic field at P is
A
$\dfrac{\mu_0 I}{16\sqrt{2}R}\left[(\sqrt{2}+1)\hat{i} + \hat{j}\right]$
B
$\dfrac{\mu_0 I}{16\sqrt{2}R}\left[\sqrt{2}\hat{i} + \hat{j}\right]$
C
$\dfrac{\mu_0 I}{16R}\left[(\sqrt{2}+1)\hat{i} + \hat{j}\right]$
D
$\dfrac{\mu_0 I}{16R}\left[\sqrt{2}\hat{i} + \hat{j}\right]$
2
KCET 2026
MCQ (Single Correct Answer)
+1
-0
Biot-Savart law indicates that an electron moving with a velocity $\vec{V}$ produces a magnetic field $\vec{B}$ around it such that
A
$\vec{B}$ is parallel to $\vec{V}$
B
$\vec{B}$ is perpendicular to $\vec{V}$
C
$\vec{B}$ is anti-parallel to $\vec{V}$
D
$\vec{B}$ is inclined to $\vec{V}$ by $45^\circ$
3
KCET 2026
MCQ (Single Correct Answer)
+1
-0
A proton, an electron and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p, R_e$ and $R_\alpha$ are the radii of circular paths of these particles, then
A
$R_\alpha = R_p = R_e$
B
$R_\alpha > R_p > R_e$
C
$R_\alpha < R_p < R_e$
D
$R_\alpha > R_p = R_e$
4
KCET 2026
MCQ (Single Correct Answer)
+1
-0
In the figure shown, the conductor PQ of length $l$ is moved from $x = 0$ to $x = b$ and then up to $x = 2b$ with a constant velocity $\vec{v}$. A uniform magnetic field $\vec{B}$ is perpendicular to the plane of the paper and extends from $x = 0$ to $x = b$ and it is zero from $x > b$. The magnitude of emf Induced in the conductor is
A
$B l x; 0 \leq x \leq b$
B
Zero; $0 \leq x \leq b$
C
$B l v; 0 \leq x \leq b$
D
$B l v; b \leq x \leq 2b$