AIPMT 2015 | Moving Charges and Magnetism Question 26 | Physics

AIPMT 2015

MCQ (Single Correct Answer)

-1

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be

1.5 MeV

1 MeV

4 MeV

0.5 MeV

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)

-1

An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude

$${{{\mu _0}{n^2}e} \over r}$$

$${{{\mu _0}ne} \over {2r}}$$

$${{{\mu _0}ne} \over {2\pi r}}$$

Zero

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)

-1

A conducting square frame of side 'a' and a long straight wire carrying current $$I$$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity 'V'. The emf induced in the frame will be proportional to

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 64 English

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 64 English

$${1 \over {{{\left( {2x + a} \right)}^2}}}$$

$${1 \over {\left( {2x - a} \right)\left( {2x + a} \right)}}$$

$${1 \over {{x^2}}}$$

$${1 \over {{{\left( {2x - a} \right)}^2}}}$$

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)

-1

A wire carrying current $$I$$ has the shape shown in adjoining figure.

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 65 English

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 65 English

Linear parts of the wire are very long and parallel to X-axis while semicircular protion of radius R is lying in Y-Z plane. Magtnetic field at pont $$O$$ is

$$\overrightarrow B = - {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right)$$

$$\overrightarrow B = {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i - 2\widehat k} \right)$$

$$\overrightarrow B = {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right)$$

$$\overrightarrow B = - {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i - 2\widehat k} \right)$$

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