1
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
The straight wire $ A B $ carries a current $ I $. The ends of the wire subtend angles $ \theta_{1} $ and $ \theta_{2} $ at the point $ P $ as shown in figure. The magnetic field at the point $ P $ is

BITSAT 2024 Physics - Moving Charges and Magnetism Question 1 English

A
$ \frac{\mu_{0} I}{4 \pi d}\left(\sin \theta_{1}-\sin \theta_{2}\right) $
B
$ \frac{\mu_{0} I}{4 \pi d}\left(\sin \theta_{1}+\sin \theta_{2}\right) $
C
$ \frac{\mu_{0} I}{4 \pi d}\left(\cos \theta_{1}-\cos \theta_{2}\right) $
D
$ \frac{\mu_{0} I}{4 \pi d}\left(\cos \theta_{1}+\cos \theta_{2}\right) $
2
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

A charged particle moving in a uniform magnetic field and losses $$16 \%$$ of its kinetic energy. The radius of curvature of its path changes by

A
8%
B
4%
C
10%
D
16%
3
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

An electric current I enters and leaves a uniform circular wire of radius r through diametrically opposite points. A charged particle q moves along the axis of circular wire passes through its centre at speed v. The magnetic force on the particle when it passes through the centre has a magnitude.

A
$${{qv{\mu _0}I} \over {2\pi r}}$$
B
$$qv{{{\mu _0}I} \over {\pi r}}$$
C
$${{qv{\mu _0}I} \over r}$$
D
0
4
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

Two particles A and B of masses mA and mB respectively, are having same charge and moving on same plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in figure. Then,

BITSAT 2021 Physics - Moving Charges and Magnetism Question 5 English

A
mA = mB and vA = vB
B
mAvA > mBvB
C
mA < mB and vA < vB
D
mAvA < mBvB
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