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Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform mag. field and describe circular path of radius $R_1$ and $R_2$ respectively. The ratio of mass of X to that of Y is .......................

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Explanation

$ r = \sqrt { 2mk \over qB } $ $ r \alpha \sqrt m \Rightarrow { m_1 \over m_2 } = \left( { R_1 \over R_2 } \right) ^2 $

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A proton and an particle are projected with the same kinetic energy at right angles to the uniform mag. field. Which one of the following statements will be true.

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Explanation

$ r = \sqrt { 2mk \over qB } $ $ r \alpha { \sqrt m \over q } $ ( k.E and B are same ) $ { rp \over r \alpha } = 1 \Rightarrow r_p = r_{alpha } $

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A 2 Mev proton is moving perpendicular to a uniform magnetic field of 2.5 tesla. The force on the proton is

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Explanation

$ F = Bq \nu $ $ = Bq \sqrt { 2E \over M } $ $ = 7.6 \times 10^{72} N$

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A charged particle moves in a uniform mag. field. The velocity of the particle at some instant makes an acute angle with the mag. field. The path of the particle will be

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Explanation

When particle entries at angles other than $ 0 ^\circ or 90 ^\circ or 180 ^\circ $ , path followed is helix

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A deutron of K. E. 50 kev is describing a circular orbit of radius 0.5 m in a plane perpendicular to magnetic field $ \vec B $ . The K.E. of the proton that describe a circular orbit of radius 0.5 min the same plane with the same $ \vec B $ is ......................

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Explanation

$ r = { \sqrt { 2m_1 Ek_1} \over Bq_1 } = {\sqrt { 2m_2 Ek_2 } \over Bq_2 } $ $ Ek_2 = { m_1 \over m_2} {q_2 \over q_1} Ek_1 $ $ = { 2m \over m } \times {q \over q} \times 50 keV$ $ = 2 \times 50 = 100 keV $

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A magenetic field existinf in a region is given by $ \vec B = Bo \left[ 1 + { x \over l } \right] \hat k $ . A squar loop of side l and carrying current I is placed with edges (sides) parallel to X-Y axis. The magnitude ofthe net magnetic force experienced by the Loop is .................

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Explanation

The mag. force (Fm) on wire ab and cd is equal and opposite, hence cancelled each other $ F_1 = F_{ad} = BoI \left[ 1 + { 90 \over l} \right] \hat k $ $ F_2 = F_{cb} = BoI \left[ 1 + { 90+l \over l} \right] \hat k $ $F_1 = F_1 - F_2 = Bo.I.l $

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The forces existing between two parallel current carrying conductors is F. If the current in each conductor is doubled, then the value of force will be

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Explanation

$ F = { \mu_o \over 2 \pi } { I_1 I_2 l \over y } $ $ F \alpha I_1 I_2 $ 4 times

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At a given place the horizontal componnent of earth's field is 0.2 G. If a vertical wire carries a current of 30 Amp upward, what is the magnitude and direction of the force on 1 meter of wire ?

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Explanation

$ F = B I l = 6 \times 10^{-4} $ east to west

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The deflection in a Galvanometer falls from 50 division to 20 when $ 12 \Omega $ shunt is applied. The Galvanometer resistance is .............................

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Explanation

I = 50 k ; $ I_G = 20 k $ where K = figure of merit $ S = {G.I_G \over I-I_G } $ $ G = 18 \Omega $

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In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an ele. potential V and then made to describe semicircular paths of radius R using a magnetic field B.If V and B are kept constant, the ratio $ { Charge on the ion \over mass of the ion }$ will be proportional to

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Explanation

ACC to work -energy theorem $ qV = {1 \over 2 } mv^2 $ ( for ele.field ) $ BqV = { mv^2 \over R } $ ( for mag . Field ) $ v = { BqR \over m } …..(2) $ sub .eq (2) in (1) $ qV = {1 \over 2 } m { B^2 q^2 R^2 \over m^2 } $ $ V = { B^2 R^2 \over 2 } { q \over m } $ $ {q \over m } = { 2V \over B^2 R^2 } $ $ {q \over m} \alpha {1 \over A^2 } $

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