Chapter 13

JEE Physics · 85 exercises

Problem 1861

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where \(\mathrm{dx}=1 \mathrm{~cm}\) ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)

6 step solution

Problem 1862

Two straight long conductors \(\mathrm{AOB}\) and \(\mathrm{COD}\) are perpendicular to each other and carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). The magnitude of the mag. field at a point " \(\mathrm{P}^{n}\) at a distance " \(\mathrm{a}^{\prime \prime}\) from the point "O" in a direction perpendicular to the plane \(\mathrm{ABCD}\) is (a) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}+I_{2}\right)\) (b) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}-I_{2}\right)\) (c) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}\) (d) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left[\left(I_{1} I_{2}\right) /\left(I_{1}-I_{2}\right)\right]\)

4 step solution

Problem 1867

If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be (a) Only inside the pipe (b) Only outside the pipe (c) Neither inside nor outside the pipe (d) Both inside and outside the pipe

4 step solution

Problem 1868

The magnetic induction at a point \(P\) which is at a distance \(4 \mathrm{~cm}\) from a long current carrying wire is \(10^{-8}\) tesla. The field of induction at a distance \(12 \mathrm{~cm}\) from the same current would be tesla. (a) \(3.33 \times 10^{-9}\) (b) \(1.11 \times 10^{-4}\) (c) \(3 \times 10^{-3}\) (d) \(9 \times 10^{-2}\)

7 step solution

Problem 1869

The strength of the magnetic field at a point \(\mathrm{y}\) near a long straight current carrying wire is \(\mathrm{B}\). The field at a distance \(\mathrm{y} / 2\) will be (a) B/2 (b) B \(/ 4\) (c) \(2 \mathrm{~B}\) (d) \(4 \mathrm{~B}\)

5 step solution

Problem 1870

The mag. field (B) at the centre of a circular coil of radius "a", through which a current I flows is (a) \(\mathrm{B} \propto \mathrm{c} \mathrm{a}\) (b) \(\mathrm{B} \propto(1 / \mathrm{I})\) (c) \(\mathrm{B} \propto \mathrm{I}\) (d) \(B \propto I^{2}\)

3 step solution

Problem 1871

A current of a \(1 \mathrm{Amp}\) is passed through a straight wire of length 2 meter. The magnetic field at a point in air at a distance of 3 meters from either end of wire and lying on the axis of wire will be (a) \(\left(\mu_{0} / 2 \pi\right)\) (b) \(\left(\mu_{0} / 4 \pi\right)\) (c) \(\left(\mu_{0} / 8 \pi\right)\) (d) zero

5 step solution

Problem 1873

A long straight wire of radius "a" carries a steady current I the current is uniformly distributed across its cross-section. The ratio of the magnetic field at \(\mathrm{a} / 2\) and \(2 \mathrm{a}\) is (a) \((1 / 4)\) (b) 4 (c) 1 (d) \((1 / 2)\)

4 step solution

Problem 1874

At a distance of \(10 \mathrm{~cm}\) from a long straight wire carrying current, the magnetic field is \(4 \times 10^{-2}\). At the distance of \(40 \mathrm{~cm}\), the magnetic field will be Tesla. (a) \(1 \times 10^{-2}\) (b) \(2 \times \overline{10^{-2}}\) (c) \(8 \times 10^{-2}\) (d) \(16 \times 10^{-2}\)

3 step solution

Problem 1876

A He nucleus makes a full rotation in a circle of radius \(0.8\) meter in \(2 \mathrm{sec}\). The value of the mag. field \(\mathrm{B}\) at the centre of the circle will be \(\quad\) Tesla. (a) \(\left(10^{-19} / \mu_{0}\right)\) (b) \(10^{-19} \mu_{0}\) (c) \(2 \times 10^{-10} \mathrm{H}_{0}\) (d) \(\left[\left(2 \times 10^{-10}\right) / \mu_{0}\right]\)

3 step solution

Problem 1879

The direction of mag. field lines close to a straight conductor carrying current will be (a) Along the length of the conductor (b) Radially outward (c) Circular in a plane perpendicular to the conductor (d) Helical

3 step solution

Problem 1880

0: Due to 10 Amp of current flowing in a circular coil of \(10 \mathrm{~cm}\) radius, the mag. field produced at its centre is \(\pi \times 10^{-3}\) Tesla. The number of turns in the coil will be (a) 5000 (b) 100 (c) 50 (d) 25

3 step solution

Problem 1883

In a H-atom, an electron moves in a circular orbit of radius \(5.2 \times 10^{-11}\) meter and produces a mag. field of \(12.56\) Tesla at its nucleus. The current produced by the motion of the electron will be (a) \(6.53 \times 10^{-3}\) (b) \(13.25 \times 10^{-10}\) (c) \(9.6 \times 10^{6}\) (d) \(1.04 \times 10^{-3}\)

3 step solution

Problem 1884

A conducting rod of 1 meter length and \(1 \mathrm{~kg}\) mass is suspended by two vertical wires through its ends. An external magnetic field of 2 Tesla is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [take \(\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\) (a) \(0.5 \mathrm{Amp}\) (b) \(15 \mathrm{Amp}\) (c) \(5 \mathrm{Amp}\) (d) \(1.5 \mathrm{Amp}\)

4 step solution

Problem 1885

A straight wire of mass \(200 \mathrm{gm}\) and length \(1.5\) meter carries a current of 2 Amp. It is suspended in mid-air by a uniform horizontal magnetic field B. [take \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]\). The \(\mathrm{B}\) is (a) \(\overline{(2 / 3) \text { tes } 1 a}\) (b) \((3 / 2)\) tesla (c) \((20 / 3)\) tesla (d) (3/20) tesla

5 step solution

Problem 1886

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current of \(2.5 \mathrm{Amp}\). The mag. field at its centre is tesla. (a) \(\pi \times 10^{-2}\) (b) \(2 \pi \times 10^{-2}\) (c) \(3 \pi \times 10^{-2}\) (d) \(4 \pi \times 10^{-2}\)

5 step solution

Problem 1887

Two concentric co-planar circular Loops of radii \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) carry currents of respectively \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) in opposite directions. The magnetic induction at the centre of the Loops is half that due to \(\mathrm{I}_{1}\) alone at the centre. If \(\mathrm{r}_{2}=2 \mathrm{r}_{1}\) the value of \(\left(\mathrm{I}_{2} / \mathrm{I}_{1}\right)\) is (a) 2 (b) \(1 / 2\) (c) \(1 / 4\) (d) 1

5 step solution

Problem 1889

For the mag. field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be (a) \(0^{\circ}\) (b) \(90^{\circ}\) (c) \(180^{\circ}\) (d) \(45^{\circ}\)

3 step solution

Problem 1891

A long straight wire carrying current of \(30 \mathrm{Amp}\) is placed in an external uniform mag. field of induction \(4 \times 10^{-4}\) tesla. The mag. field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point \(2 \mathrm{~cm}\) away from the wire is tesla. (a) \(10^{-4}\) (c) \(5 \times 10^{-4}\) (b) \(3 \times 10^{-4}\) (d) \(6 \times 10^{-4}\)

4 step solution

Problem 1892

Two similar coils are kept mutually perpendicular such that their centers co- inside. At the centre, find the ratio of the mag. field due to one coil and the resultant magnetic field by both coils, if the same current is flown. (a) \(1: \sqrt{2}\) (b) \(1: 2\) (c) \(2: 1\) (d) \(\sqrt{3}: 1\)

3 step solution

Problem 1893

A long wire carr1es a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is \(\mathrm{B}\). It is then bent into a circular Loop of n turns. The magnetic field at the centre of the coil for same current will be. (a) \(\mathrm{nB}\) (b) \(\mathrm{n}^{2} \mathrm{~B}\) (c) \(2 \mathrm{nB}\) (d) \(2 \mathrm{n}^{2} \mathrm{~B}\)

3 step solution

Problem 1894

The mag. field due to a current carrying circular Loop of radius \(3 \mathrm{~cm}\) at a point on the axis at a distance of \(4 \mathrm{~cm}\) from the centre is \(54 \mu \mathrm{T}\) what will be its value at the centre of the LOOP. (a) \(250 \mu \mathrm{T}\) (b) \(150 \mu \mathrm{T}\) (c) \(125 \mu \mathrm{T}\) (d) \(75 \mu \mathrm{T}\)

4 step solution

Problem 1895

5: When the current flowing in a circular coil is doubled and the number of turns of the coil in it is halved, the magnetic field at its centre will become (a) Four times (b) Same (c) Half (d) Double

6 step solution

Problem 1896

Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moment is (a) \(2: \pi\) (b) \(\pi: 2\) (c) \(\pi: 4\) (d) \(4: \pi\)

3 step solution

Problem 1897

Two concentric coils each of radius equal to \(2 \pi \mathrm{cm}\) are placed at right angles to each other. 3 Amp and \(4 \mathrm{Amp}\) are the currents flowing in each coil respectively. The magnetic field intensity at the centre of the coils will be Tesla. (a) \(5 \times 10^{-5}\) (b) \(7 \times 10^{-5}\) (c) \(12 \times 10^{-5}\) (d) \(10^{-5}\)

3 step solution

Problem 1898

Two parallel long wires \(\mathrm{A}\) and B carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). \(\left(\mathrm{I}_{2}<\mathrm{I}_{1}\right)\) when \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) are in the same direction the mag. field at a point mid way between the wires is \(10 \mu \mathrm{T}\). If \(\mathrm{I}_{2}\) is reversed, the field becomes \(30 \mu \mathrm{T}\). The ratio \(\left(\mathrm{I}_{1} / \mathrm{I}_{2}\right)\) is (a) 1 (b) 2 (c) 3 (d) 4

5 step solution

Problem 1901

A solenoid of \(1.5\) meter length and \(4 \mathrm{~cm}\) diameter possesses 10 turn per \(\mathrm{cm}\). A current of \(5 \mathrm{Amp}\) is flowing through it. The magnetic induction at axis inside the solenoid is (a) \(2 \pi \times 10^{-3} \mathrm{~T}\) (b) \(2 \pi \times 10^{-5} \mathrm{~T}\) (c) \(2 \pi \times 10^{-2} \mathrm{G}\) (d) \(2 \pi \times 10^{-5} \mathrm{G}\)

4 step solution

Problem 1902

A straight wire of length \(30 \mathrm{~cm}\) and mass 60 milligram lies in a direction \(30^{\circ}\) east of north. The earth's magnetic field at this site is horizontal and has a magnitude of \(0.8 \mathrm{G}\). What current must be passed through the wire so that it may float in air ? \(\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (a) \(10 \mathrm{Amp}\) (b) \(20 \mathrm{Amp}\) (c) \(40 \mathrm{Amp}\) (d) \(50 \mathrm{Amp}\)

3 step solution

Problem 1903

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of \(50 \mathrm{Amp}\). It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

4 step solution

Problem 1906

Two particles \(\mathrm{X}\) and \(\mathrm{Y}\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform mag. field and describe circular path of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) respectively. The ratio of mass of \(\mathrm{X}\) to that of \(\mathrm{Y}\) is (a) \(\sqrt{\left(r_{1} / \mathrm{r}_{2}\right)}\) (b) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (c) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (b) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\)

3 step solution

Problem 1907

An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge \(1.6 \times 10^{-19} \mathrm{C}\) and moving with a velocity of \(10^{6} \mathrm{~m} / \mathrm{s}\) enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)

5 step solution

Problem 1908

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. (a) The \(\alpha\) - particle will be bent in a circular path with a small radius that for the proton. (b) The radius of the path of the \(\alpha\) - particle will be greater than that of the proton. (c) The \(\alpha\) - particle and the proton will be bent in a circular path with the same radius. (d) The \(\alpha\) - particle and the proton will go through the field in a straight line.

5 step solution

Problem 1909

A 2 Mev proton is moving perpendicular to a uniform magnetic field of \(2.5\) tesla. The force on the proton is (a) \(3 \times 10^{-10} \mathrm{~N}\) (b) \(70.8 \times 10^{-11} \mathrm{~N}\) (c) \(3 \times 10^{-11} \mathrm{~N}\) (d) \(7.68 \times 10^{-12} \mathrm{~N}\)

3 step solution

Problem 1910

A proton is projected with a speed of \(2 \times 10^{6}(\mathrm{~m} / \mathrm{s})\) at an angle of \(60^{\circ}\) to the \(\mathrm{X}\) -axis. If a uniform mag. field of \(0.104\) tesla is applied along \(\mathrm{Y}\) -axis, the path of proton is (a) A circle of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period \(\pi \times 10^{-7} \mathrm{sec}\) (b) A circle of \(\mathrm{r}=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{sec}\) (c) A helix of \(r=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{sec}\) (d) A helix of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period \(4 \pi \times 10^{-7} \mathrm{sec}\)

4 step solution

Problem 1911

1: 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 (a) A straight line (b) A circle (c) A helix with uniform pitch (d) A helix with non-uniform pitch

5 step solution

Problem 1916

A conducting rod of length \(\ell\) [cross-section is shown] and mass \(\mathrm{m}\) is moving down on a smooth inclined plane of inclination \(\theta\) with constant speed v. A vertically upward mag. field \(\mathrm{B}^{-}\) exists in upward direction. The magnitude of mag. field \(B^{-}\) is(a) \([(\mathrm{mg} \sin \theta) /(\mathrm{I} \ell)]\) (b) \([(\mathrm{mg} \cos \theta) /(\mathrm{I} \ell)]\) (c) \([(\mathrm{mg} \tan \theta) /(\mathrm{I} \ell)]\) (d) \([(\mathrm{mg}) /(\mathrm{I} \ell \sin \theta)]\)

4 step solution

Problem 1918

A magnetic field existing in a region is given by \(\mathrm{B}^{-}=\mathrm{B}_{0}[1+(\mathrm{x} / \ell)] \mathrm{k} \wedge . \mathrm{A}\) square loop of side \(\ell\) and carrying current I is placed with edges (sides) parallel to \(\mathrm{X}-\mathrm{Y}\) axis. The magnitude of the net magnetic force experienced by the Loop is (a) \(2 \mathrm{~B}_{0} \overline{\mathrm{I} \ell}\) (b) \((1 / 2) B_{0} I \ell\) (c) \(\mathrm{B}_{\circ} \mathrm{I} \ell\) (d) BI\ell

3 step solution

Problem 1919

The forces existing between two parallel current carrying conductors is \(\mathrm{F}\). If the current in each conductor is doubled, then the value of force will be (a) \(2 \mathrm{~F}\) (b) \(4 \mathrm{~F}\) (c) \(5 \mathrm{~F}\) (d) \((\mathrm{F} / 2)\)

5 step solution

Problem 1921

A Galvanometer has a resistance \(\mathrm{G}\) and \(\mathrm{Q}\) current \(\mathrm{I}_{\mathrm{G}}\) flowing in it produces full scale deflection. \(\mathrm{S}_{1}\) is the value of the shunt which converts it into an ammeter of range 0 to \(\mathrm{I}\) and \(\mathrm{S}_{2}\) is the value of the shunt for the range 0 to \(2 \mathrm{I}\). The ratio \(\left(\mathrm{S}_{1} / \mathrm{S}_{2}\right) \mathrm{is}\) (a) \(\left[\left(2 \mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) /\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)\right]\) (b) \((1 / 2)\left[\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) /\left(2 \mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)\right]\) (c) 2 (d) 1

3 step solution

Problem 1923

The deflection in a Galvanometer falls from 50 division to 20 when \(12 \Omega\) shunt is applied. The Galvanometer resistance is (a) \(18 \Omega\) (b) \(36 \Omega\) (c) \(24 \Omega\) (d) \(30 \Omega\)

5 step solution

Problem 1924

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an ele. potential \(\mathrm{V}\) and then made to describe semicircular paths of radius \(\mathrm{r}\) using a magnetic field \(\mathrm{B}\). If \(\mathrm{V}\) and \(\mathrm{B}\) are kept constant, the ratio [(Charge on the ion) / (mass of the ion)] will be proportional to. (a) \(\left(1 / r^{2}\right)\) (b) \(r^{2}\) (c) \(\mathrm{r}\) (d) \((1 / \mathrm{r})\)

5 step solution

Problem 1925

A Galvanometer of resistance \(15 \Omega\) is connected to a battery of 3 volt along with a resistance of \(2950 \Omega\) in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (a) \(6050 \Omega\) (b) \(4450 \Omega\) (c) \(5050 \Omega\) (d) \(5550 \Omega\)

3 step solution

Problem 1926

A Galvanometer coil has a resistance of \(15 \Omega\) and gives full scale deflection for a current of \(4 \mathrm{~mA}\). To convert it to an ammeter of range 0 to \(6 \mathrm{Amp}\) (a) \(10 \mathrm{~m} \Omega\) resistance is to be connected in parallel to the galvanometer. (b) \(10 \mathrm{~m} \Omega\) resistance is to be connected in series with the galvanometer. (c) \(0.1 \Omega\) resistance is to be connected in parallel to the galvanometer. (d) \(0.1 \Omega\) resistance is to be connected in series with the galvanometer.

6 step solution

Problem 1927

The deflection in moving coil Galvanometer is reduced to half when it is shunted with a \(40 \Omega\) coil. The resistance of the Galvanometer is (c) \(40 \Omega\) (a) \(60 \Omega\) (b) \(\overline{10 \Omega}\) (d) \(20 \Omega\)

3 step solution

Problem 1929

A conducting circular loop of radius a carries a constant current I. It is placed in a uniform magnetic field \(\mathrm{B}^{-}\), such that \(\mathrm{B}^{-}\) is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is (a) \(\mathrm{B}^{-} \operatorname{Ir}\) (b) \(\mathrm{B}^{-} \mathrm{I} \pi \mathrm{r}^{2}\) (c) Zero (d) BI \((2 \pi \mathrm{r})\)

3 step solution

Problem 1930

Two thin long parallel wires separated by a distance \(\mathrm{y}\) are carrying a current I Amp each. The magnitude of the force per unit length exerted by one wire on other is (a) \(\left[\left(\mu_{0} I^{2}\right) / y^{2}\right]\) (b) \(\left[\left(\mu_{o} I^{2}\right) /(2 \pi \mathrm{y})\right]\) (c) \(\left[\left(\mu_{0}\right) /(2 \pi)\right](1 / y)\) (d) \(\left[\left(\mu_{0}\right) /(2 \pi)\right]\left(1 / \mathrm{y}^{2}\right)\)

3 step solution

Problem 1932

If two streams of protons move parallel to each other in the same direction, then they (a) Do not exert any force on each other (b) Repel each other (c) Attract each other (d) Get rotated to be perpendicular to each other.

4 step solution

Problem 1933

A coil in the shape of an equilateral triangle of side 115 suspended between the pole pieces of a permanent magnet such that \(\mathrm{B}^{-}\) is in plane of the coil. If due to a current \(\mathrm{I}\) in the triangle a torque \(\tau\) acts on it, the side 1 of the triangle is (a) \((2 / \sqrt{3})(\tau / \mathrm{BI})^{1 / 2}\) (b) \((2 / 3)(\tau / B I)\) (c) \(2[\tau /\\{\sqrt{(} 3) \mathrm{BI}\\}]^{1 / 2}\) (d) \((1 / \sqrt{3})(\tau / \mathrm{BI})\)

4 step solution

Problem 1935

The unit of ele. current "AMPERE" is the current which when flowing through each of two parallel wires spaced 1 meter apart in vacuum and of infinite length will give rise to a force between them equal to \(\mathrm{N} / \mathrm{m}\) (a) 1 (b) \(2 \times 10^{-7}\) (c) \(1 \times 10^{-2}\) (d) \(4 \pi \times 10^{-7}\)

4 step solution

Problem 1938

A particle of mass \(\mathrm{m}\) and charge q moves with a constant velocity v along the positive \(\mathrm{x}\) -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 minimum value of required so that the particle can just enter the region \(\mathrm{x}>\mathrm{b}\) is (a) \([(\mathrm{q} \mathrm{bB}) / \mathrm{m}]\) (b) \(q(b-a)(B / m)\) (c) \([(\mathrm{qaB}) / \mathrm{m}]\) (d) \(q(b+a)(B / 2 m)\)

5 step solution

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