MHT CET 2024 10th May Evening Shift | Moving Charges and Magnetism Question 31 | Physics

A charged particle is moving in a uniform magnetic field in a circular path of radius ' $R$ '. When the kinetic energy of the particle is increased to three times, then the new radius will be

$\frac{\mathrm{R}}{3}$

$\frac{\mathrm{R}}{\sqrt{3}}$

$\sqrt{3} \cdot \mathrm{R}$

$3 \cdot \mathrm{R}$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

-0

A circular arc of radius r carrying current ' I ' subtends an angle $\frac{\pi}{8}$ at its entre. The radius of a metal wire is uniform. The magnetic induction at the centre of circular arc is ( $\mu_0=$ permeability of free space)

$\frac{\mu_0 \mathrm{I}}{8 \mathrm{r}}$

$\frac{\mu_0 I}{32 r}$

$\frac{\mu_0 I}{64 r}$

$\frac{\mu_0 I}{16 r}$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

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Electron of mass ' $m$ ' and charge ' $q$ ' is travelling with speed ' $v$ ' along a circular path of radius ' $R$ ' at right angles to a uniform magnetic field of intensity ' B '. If the speed of the electron is halved and the magnetic field is doubled, the resulting path would have radius

$\frac{\mathrm{R}}{2}$

$\frac{\mathrm{R}}{4}$

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

-0

A current carrying circular loop of radius ' $R$ ' and current carrying long straight wire are placed in the same plane. $I_c$ and $I_w$ are the currents through circular loop and long straight wire respectively. The perpendicular distance between centre of the circular loop and wire is ' d '. The magnetic field at the centre of the loop will be zero when separation ' $d$ ' is equal to

$\frac{\mathrm{RI}_w}{\pi \mathrm{I}_{\mathrm{c}}}$

$\frac{\mathrm{RI}_{\mathrm{c}}}{\pi \mathrm{I}_{\mathrm{w}}}$

$\frac{\pi \mathrm{I}_{\mathrm{c}}}{R \mathrm{I}_{\mathrm{w}}}$

$\frac{\pi \mathrm{I}_{\mathrm{w}}}{\mathrm{R}_{\mathrm{c}}}$

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