AIPMT 2015 Cancelled Paper | Moving Charges and Magnetism Question 33 | Physics

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 70 English

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 70 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 71 English

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 71 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

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

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

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

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

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 2014

MCQ (Single Correct Answer)

-1

Two identical long conducting wires $$AOB$$ and $$COD$$ are placed at right angle to each other, with one above other such that $$O$$ is their common point for the two. The wires carry $$I$$₁ and $$I$$₂ currents, respectively. Point $$P$$ is lying at distance f from $$O$$ along a direction perpendicular to the plane containing the wires. The magnetic field at the point $$P$$ will be

$${{{\mu _0}} \over {2\pi d}}\left( {{{{I_1}} \over {{I_2}}}} \right)$$

$${{{\mu _0}} \over {2\pi d}}\left( {{I_1} + {I_2}} \right)$$

$${{{\mu _0}} \over {2\pi d}}\left( {I_1^2 - I_2^2} \right)$$

$${{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{1/2}}$$

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