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NEET Practice Questions (MCQs) with Answers & Solutions

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A particle of mass mand charge q moves witha constant velocity $ \nu $ along the positive x- direction. It enters a region containing a uniform magnetic field B directed along the negative z-direction, extending from x= a to x= b. The minimumvalue of required so that the particle can just enter the region $ x \gt b $ is

A. $ { q bB \over m } $ B. $ q (b-a ) { B \over m } $ C. $ { q a B \over m } $ D. $ q (b+a ) { B \over 2m } $

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Explanation

In the fig. the z-axis points outof the paper and mag. field is direced into the paper represented by It is present between PQ and RS only The particle moves in a circular path of radius r in the magnetic field. It can just enter the region $x \gt 9 $ for $ Now \;r = { mv \over qB } \geq (b-a) $ $ \nu \geq { (b-a) qB \over m } $ $ \therefore \nu min = { qB ( b -a ) \over n } $

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An iron rod of length L and magnetic moment M is bent in the form of a semi circle. Now its magnetic moment will be

A. M B. $ { 2M \over \pi } $ C. $ { M \over \pi } $ D. $ M \pi $

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Explanation

On bending a rod its pole strangth remains unchanged where as its magnetic moment changes. M' = m (2 R ) $ = m \left( 2 { L \over \pi } \right) = { 2 \over \pi } mL = { 2m \over \pi } $

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Unit of magnetic Flux density is ...........................

A. Tesla B. $ { Weber \over meter^2 } $ C. $ { Newton \over Amp- meter} $ D. All of the above

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Explanation

All of the above

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Magnetic intensity for an axial point due to a short bar magnet of magnetic moment M is given by

A. $ { \mu_o \over 4 \pi } { M \over d^3 } $ B. $ { \mu_o \over 4 \pi } { M \over d^2 } $ C. $ { \mu_o \over 2 \pi } { M \over d^3 } $ D. $ { \mu_o \over 2 \pi } { M \over d^2 } $

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A magnet of magnetic moment M and pole strength m is divided in two equal parts, then magnetic moment of each part will be

A. M B. $ { M \over 2 } $ C. $ { M \over 4 } $ D. 2M

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Explanation

Case (i) If cut along the axis of magnet of length l , then new pole strength m' = m/2 and new length l ' = l $ \therefore New magnetic moment M' = { m \over 2 } \times l = { ml \over 2 } = { M \over 2 } $ Case (ii) If cut perpendicular to the axis of magnet, then new pole strength m' = m
and new length l' = l /2 $ \therefore New magnetic moment M' = { m \over 2 } \times l = { ml \over 2 } = { M \over 2 } $

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If a magnet of pole strengthm is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be

A. m/4 B. m/2 C. m/8 D. 4m

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Explanation

For each part m' = m/2

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The magnetism of magnet is due to

A. The spin motion of electron B. Earth C. Pressure inside the earth core region D. Cosmic rays

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Explanation

The spin motion of electron

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The magnetic field at a point x on the axis of a small bar magnet is equal to the at a point y on the equator of the same magnet. The ratio of the distances of x and y from the centre of the magnet is

A. $ 2 ^ {-3 } $ B. $ 2 ^ {-1/3 } $ C. $ 2 ^ {3 } $ D. $ 2 ^ {+1/3 } $

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Explanation

$ on the axis B_1 = { 2m \over x^3 } $ $ on the equator B_2 = { M \over y^3 } $ $ As B_1 = B_2 $ $ { 2M \over x^3 } = { M \over y^3 } $ $ { x^3 \over y^3 } = 2 $ $ { x \over y } = 2 ^ { 1 \over 3 } $

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The magnetic field due to a short magnet at a point on its axis at a distance x cm from the middle point of the magnet is 200 gauss. The magnetic field at a point on the neutral axis at a distance xcm from the middle of the magnet is gauss.

A. 100 B. 400 C. 50 D. 200

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Explanation

$ on the axis B axis = { 2m \over x^3 } $ $ 200 gauss = { 2M \over x^3 } $ $ 100 gauss= { M \over x^3 } $ 100 gauss = Bequator

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A bar magnet having a magnetic moment of $ 2 \times 10^ 4 J T^{-1} $ is free to rotate in a horizontal plane. A horizontal magnetic field $ B = 6 \times 10^{-4} $ Tesla exists in the space. The workdone in taking the magnet slowly from a direction parallel to the field to a direction $60^ \circ $ from the field is

A. 0.6J B. 12J C. 6J D. 2J

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Explanation

$ W = MB( 1 -cos \theta ) $ $ = 2 \times 10^4 \times 6 \times 10^{-4} (1 - cos 60 ^\circ) $ = 6 joule

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