KCET 2026 | Moving Charges and Magnetism Question 1 | Physics

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

$\dfrac{\mu_0 I}{16\sqrt{2}R}\left[(\sqrt{2}+1)\hat{i} + \hat{j}\right]$

$\dfrac{\mu_0 I}{16\sqrt{2}R}\left[\sqrt{2}\hat{i} + \hat{j}\right]$

$\dfrac{\mu_0 I}{16R}\left[(\sqrt{2}+1)\hat{i} + \hat{j}\right]$

$\dfrac{\mu_0 I}{16R}\left[\sqrt{2}\hat{i} + \hat{j}\right]$

KCET 2026

MCQ (Single Correct Answer)

-0

Biot-Savart law indicates that an electron moving with a velocity $\vec{V}$ produces a magnetic field $\vec{B}$ around it such that

$\vec{B}$ is parallel to $\vec{V}$

$\vec{B}$ is perpendicular to $\vec{V}$

$\vec{B}$ is anti-parallel to $\vec{V}$

$\vec{B}$ is inclined to $\vec{V}$ by $45^\circ$

KCET 2026

MCQ (Single Correct Answer)

-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

$R_\alpha = R_p = R_e$

$R_\alpha > R_p > R_e$

$R_\alpha < R_p < R_e$

$R_\alpha > R_p = R_e$

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