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NEET NEET Questions — Free MCQ Practice

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A metallic loop is placed in a magnetic field. If a current is passed through it, then

A.

The ring will feel a force of attraction

B.

The ring will feel a force of repulsion

C.

It will move to and fro about its centre of gravity

D.

None of these

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Explanation

$(d) T h e r i n g w i l l f e e l a c o u p l e o f f o r c e p r o d u c i n g a n e t t o r q u e o n t h e r i n g . T h e f o r c e s a r e i n o p p o s i t e d i r e c t i o n s, s o t h e n e t f o r c e i s z e r o .$

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The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction

A.

$\frac{3}{2} \frac{h^{2}}{r^{2}}$

B.

$\frac{2}{3} \frac{h^{2}}{r^{2}}$

C.

$\frac{3}{2} \frac{r^{2}}{h^{2}}$

D.

$\frac{2}{3} \frac{r^{2}}{h^{2}}$

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Explanation

(a) Field at the centre $B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 πin}{r} = \frac{μ_{0}}{2} \cdot \frac{ni}{r}$

Field at a distance h from the centre

$B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{2 {πnir}^{2}}{(r^{2} + h^{2})^{3 / 2}} = \frac{μ_{0}}{2} \cdot \frac{{nir}^{2}}{r^{3} {(1 + \frac{h^{2}}{r^{2}})}^{3 / 2}} = B_{1} {(1 + \frac{h^{2}}{r^{2}})}^{- 3 / 2} = B_{1} (1 - \frac{3}{2} \cdot \frac{h^{2}}{r^{2}})$

(By binomial theorem)

Hence $B_{2}$ is less than $B_{1}$ by a fraction $= \frac{3}{2} \frac{h^{2}}{r^{2}}$

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In hydrogen atom, the electron is making $6.6 \times 10^{15} r e v / s e c$ around the nucleus in an orbit of radius 0.528 Å. The magnetic moment $(A - m^{2})$ will be

A.

$1 \times 10^{- 15}$

B.

$1 \times 10^{- 10}$

C.

$1 \times 10^{- 23}$

D.

$1 \times 10^{- 27}$

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Explanation

(c) $i = 6.6 \times 10^{15} \times 1.6 \times 10^{- 19} = 10.5 \times 10^{- 4} a m p$

$A = {πR}^{2} = 3.142 \times {(0.528)}^{2} \times 10^{- 2} m^{2}$

$\Rightarrow M = i A = 10.5 \times 10^{- 4} \times 3.142 \times {(0.528)}^{2} \times 10^{- 20}$

$= 10 \times 10^{- 24} u n i t s = 1 \times 10^{- 23} u n i t s$

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A proton of mass $1.67 \times 10^{- 27} k g$ and charge $1.6 \times 10^{- 19} C$ is projected with a speed of $2 \times 10^{6} m / s$ at an angle of $60 °$ to the X-axis. If a uniform magnetic field of 0.104 Tesla is applied along Y-axis, the path of the proton is:

A.

A circle of radius = 0.2 m and time period $π \times 10^{- 7} s$

B.

A circle of radius = 0.1 m and time period $2 π \times 10^{- 7} s$

C.

A helix of radius = 0.1 m and time period $2 π \times 10^{- 7} s$

D.

A helix of radius = 0.2 m and time period $4 π \times 10^{- 7} s$

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The charge on a particle Y is double the charge on particle X. These two particles X and Y after being accelerated through the same potential difference enter a region of the uniform magnetic field and describe circular paths of radii $R_{1}$ and $R_{2}$ respectively. The ratio of the mass of X to that of Y is:

A.

${(\frac{2 R_{1}}{R_{2}})}^{2}$

B.

${(\frac{R_{1}}{2 R_{2}})}^{2}$

C.

$\frac{R_{1}^{2}}{2 R_{2}^{2}}$

D.

$\frac{2 R_{1}}{R_{2}}$

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Explanation

(c)

$R = \frac{m v}{q B} = \frac{1}{B} \times \sqrt{\frac{2 m V}{q}} \frac{R_{1}}{R_{2}} = \sqrt{\frac{m_{1} q_{2}}{m_{2} q_{1}}} = \sqrt{\frac{m_{1} \times 2 q}{m_{2} q}} \frac{m_{1}}{m_{2}} = \frac{{R_{1}}^{2}}{2 {R_{2}}^{2}}$

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An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_{e}$ , $R_{p}$ , $R_{d}$ and $R_{α}$ . It follows that

A.

$R_{e} = R_{p}$

B.

$R_{p} = R_{d}$

C.

$R_{d} = R_{α}$

D.

$R_{p} = R_{α}$

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Explanation

(c) By using

$r = \frac{m v}{q B} = \frac{v}{(\frac{q}{m}) B} \Rightarrow r \propto \frac{1}{(q / m)}$

$∵ {(\frac{q}{m})}_{e^{-}} > {(\frac{q}{m})}_{p^{+}} > \{{(\frac{q}{m})}_{d} = {(\frac{q}{m})}_{α}\}$

$∵ R_{d} = R_{α}$

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An electric field of 1500 V / m and a magnetic field of 0.40 weber / $m e t e r^{2}$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is

A.

$1.6 \times 10^{15} m / s$

B.

$6 \times 10^{- 16} m / s$

C.

$3.75 \times 10^{3} m / s$

D.

$3.75 \times 10^{2} m / s$

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Explanation

(c) When electron moves in both electric and magnectic field then $q E = q v B .$

$∴ v = \frac{E}{B} = \frac{1500}{0.04} = 3750 m / s = 3.75 \times 10^{3} m / s .$

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An electron, moving in a uniform magnetic field of induction of intensity $\overset{⇀}{B,}$ has its radius directly proportional to :

A.

Its charge

B.

Magnetic field

C.

Speed

D.

None of these

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Explanation

(c) $r = \frac{m v}{q B} \Rightarrow r \propto v$

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A long straight wire along the z-axis carries a current I in the negative z-direction. The magnetic field vector $\vec{B}$ at a point having coordinates (x, y) in the z = 0 plane is :

A.

$\frac{μ_{0} I (y \hat{i} - x \hat{j})}{2 π (x^{2} + y^{2})}$

B.

$\frac{μ_{0} I (x \hat{i} + y \hat{j})}{2 π (x^{2} + y^{2})}$

C.

$\frac{μ_{0} I (x \hat{j} - y \hat{i})}{2 π (x^{2} + y^{2})}$

D.

$\frac{μ_{0} I (x \hat{i} - y \hat{j})}{2 π (x^{2} + y^{2})}$

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A particle of charge q and mass m moves in a circular orbit of radius r with angular speed $ω$ . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

A.

$ω a n d q$

B.

$ω q a n d m$

C.

$q a n d m$

D.

$ω a n d m$

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Explanation

(c) The effective current $i = \frac{q ω}{2 π} a n d A = {πr}^{2}$ .
Magnetic moment $M = i A = \frac{1}{2} q ω r^{2}$
Angular moment $L = I ω = m r^{2} ω \Rightarrow \frac{M}{L} = \frac{q}{2 m}$

NEET Practice Questions (MCQs) with Answers & Solutions

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