BITSAT 2024 | Moving Charges and Magnetism Question 1 | Physics

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

$ \frac{\mu_{0} I}{4 \pi d}\left(\sin \theta_{1}-\sin \theta_{2}\right) $

$ \frac{\mu_{0} I}{4 \pi d}\left(\sin \theta_{1}+\sin \theta_{2}\right) $

$ \frac{\mu_{0} I}{4 \pi d}\left(\cos \theta_{1}-\cos \theta_{2}\right) $

$ \frac{\mu_{0} I}{4 \pi d}\left(\cos \theta_{1}+\cos \theta_{2}\right) $

BITSAT 2023

MCQ (Single Correct Answer)

-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

10%

16%

BITSAT 2022

MCQ (Single Correct Answer)

-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.

$${{qv{\mu _0}I} \over {2\pi r}}$$

$$qv{{{\mu _0}I} \over {\pi r}}$$

$${{qv{\mu _0}I} \over r}$$

BITSAT 2021

MCQ (Single Correct Answer)

-1

Two particles A and B of masses m_A and m_B 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 v_A and v_B respectively and the trajectories are as shown in figure. Then,

BITSAT 2021 Physics - Moving Charges and Magnetism Question 5 English

m_A = m_B and v_A = v_B

m_Av_A > m_Bv_B

m_A < m_B and v_A < v_B

m_Av_A < m_Bv_B

Questions Asked from MCQ (Single Correct Answer)

BITSAT 2024 (1) BITSAT 2023 (1) BITSAT 2022 (1) BITSAT 2021 (1) BITSAT 2020 (1)

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