Q: The magnetic force acting on a charged particle of charge\$ -2\mu C\$ in a magnetic field of 2T acting in y-direction, when the particle velocity is \$(2\check{i} + 3\check{j})\$ \$\times 10^6 ms^{-1}\$ is :

Force on a moving charge in magnetic field \$ = q(\vec v \times B)\$\$ \Rightarrow \vec F =-2 \times 10^{-6} (2\hat i +3\hat j)\times 2\hat j \times 10^6\$\$ \Rightarrow \vec F =-2 \times 10^{-6} (2 \hat i\times 2\hat j) \times 10^6\$\$ \Rightarrow \vec F =-8\hat k N\$Therefore, A is correct option.

Q: A particle of mass m and charge q has an initial velocity \$\vec{v} = v_0 \hat{j}\$. If an electric field \$\vec{E} = E_0 \hat{i}\$ and magnetic field \$\vec{B} = B_0 \hat{i}\$ act on the particle, its speed will double after a time.

Given \$\vec{V}_i = V_0 \hat{j} , \ \vec{E} = E_0 \hat{i} , \vec{B} = B_0 \hat{i}\$Now we see that the velocity is perpendicular to the magnetic field so the magnitude of velocity will not change due to magnetic field.The magnitude of velocity will change in 'i' direction due to \$\vec{E} \$ only\$acc = \dfrac{force}{mass} \$ due to \$\vec{E}\$ on particle \$= \dfrac{q E_0}{m} \hat{i}\$Let us say that the speed of particle doubles after time 't'. in y direction the speed will be \$V_0\$ only in \$y - z\$ plane always perpendicular to x-direction so net speed at time 't'.\$(V_{net})^2 = \underbrace { (V_0)^2 }_{ V \ in \ y - z \ plane } + \underbrace { \left(\dfrac{q E_0}{m} t\right)^2 }_{ V_x \ after \ time\ t \, in\ x-direction\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!} \$ .Also by given condition \$V_{net} = 2V_0\$\$\Rightarrow (2 V_0)^2 = V_0^2 + \left(\dfrac{q E_0}{m} t\right)^2\$\$\Rightarrow 4 V_0^2 - V_0^2 = \left(\dfrac{q E_0}{m} t \right)^2\$\$\Rightarrow 3 V_0^2 = \left(\dfrac{q E_0}{m} t \right)^2\$\$\Rightarrow t = \dfrac{\sqrt{3} V_0 m}{q E_0}\$ Option (A) is correct.

Q: A bullet of mass \$2g\$ is with charge of \$2\mu C\$. To attain speed of \$10ms^{-1}\$,it must be accelerated through potential difference of

Velocity \$=\sqrt{\dfrac{2qV}{m}}\$\$V=\dfrac{mv^2}{2q}\$\$V=\dfrac{2\times 10^{-3}\times 10\times 10}{2\times 2\times 10^{-6}}\$\$V=50000V\$\$V=50kV\$

Q: An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?

Hint:- If a charge \$q\$ is moving with velocity \$v\$ in a Magnetic Field \$B\$, such that angle between velocity of charge and Magnetic Field is \$\theta\$. Then Force acting on charge is given by,\$F = qvBsin\theta\$Step 1: Note the given values and calculate force on an electron in magnetic field.\$\bullet\$ It is given that an electron is projected with uniform velocity along the axis of current carrying long solenoid. Let charge on electron is \$e\$ and uniform velocity of electron is \$v\$. Also let at any point on axis of solenoid Magnetic Field is \$B\$.\$\bullet\$ We know that Magnetic Field on axis of solenoid is always parallel to the axis. And since electron is moving along axis of solenoid, we have \$\theta = 0°\$ or \$\theta = 180°\$. In both the cases value of \$\sin\theta\$ is zero. \$\bullet\$ Thus net force acting on electron, that is \$F = qvBsin\theta = 0\$Step 2: Explanation for options:-\$\bullet\$ Since net force on electron is zero, by Newton's Second Law its acceleration will be zero. Option A is incorrect.\$\bullet\$ Also since net force on electron is zero, its direction of motion will not change. So Option B is incorrect.\$\bullet\$As theta is zero not 45 so option C is incorrect.\$\bullet\$ As net force is zero so net acceleration is zero. So the electron will continue to move along the axis with same velocity.Thus, only Option D is correct.

Q: A charged particle is projected in a magnetic field \$\vec{B}=(3\vec{i}+4\vec{j})\times 10^{-2}T\$ and the acceleration is found to be \$\vec{a}=(x\vec{i}+2\vec{j})m/s^{2}\$. The value of x is :

\$\overline{F}=q(\overline V\times\overline B)\$So \$\overline F\bot\overline B\$\$\therefore \overline F\cdot \overline B=0\$\$\overline ma\cdot\overline B=0\$\$\overline a\cdot\overline B=0\$\$(x\overline i+2\overline j)\cdot(3\overline i+4\overline j)=0\$\$3x+8=0\$\$x=\dfrac{-8}{3}\$

Q: Write the expression for the force on a charge moving in a magnetic field.Use this expression to define the \$SI\$ unit of magnetic field.

Force on a charge \$(q)\$ moving in a magnetic field \$B\$ with velocity \$\overrightarrow { v } \$ making an angle \$\theta\$ (with the direction of magnetic field (\$\overrightarrow { B}\$ )is given by\$F_m=qvB \sin \theta\$When \$\theta=90^o \Rightarrow \sin \theta =1\$, so \$F_m=qvB\$or \$B=\dfrac{F_m}{qv}\$If \$v=1\ m/s, B=\dfrac{F_m}{q}\ newton/coulomb.\$SI unit of magnetic field is tesla.Thus, \$1\$ tesla is the magnetic field in which a charge particle moving with velocity \$1\ m/s\$ perpendicular to velocity experiences a force of \$1 \ newton/coulomb\$.

Q: Two long and parallel straight wires carrying currents of \$2\ A\$ and \$5\$ A in the opposite directions are separated by a distance of \$1\ cm\$. Find the nature and magnitude of the magnetic force between them.

\$I_1=2 A, I_2=5A, a=1\ cm= 1 \times 10^{-2} \ m\$Force between two parallel wire per unit length is given by-\$F = \dfrac{\mu _0}{2 \pi}. \dfrac{I_1 I_2}{a} = 2 \times 10^{-7} \times \dfrac{2 \times 5}{1 \times 10^{-2}} = 20 \times 10^{-5}\ N\$ (Repulsive).Both will repel because current in both wires is opposite to each other.

Question 1

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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Question 2

A charged particle moves through a magnetic field perpendicular to its direction. Then

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When a charge particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant). Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by $\frac{1}{2} mv^2$ and $v^2$ is the square of the magnitude of velocity which does not change.In this question, we could not infer that where the momentum mentioned here is tangential or angular. If it is angular momentum, then (d) is the correct option.

Question 3

The magnetic force acting on a charged particle of charge$ -2\mu C$ in a magnetic field of 2T acting in y-direction, when the particle velocity is $(2\check{i} + 3\check{j})$ $\times 10^6 ms^{-1}$ is :

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Force on a moving charge in magnetic field $ = q(\vec v \times B)$$ \Rightarrow \vec F =-2 \times 10^{-6} (2\hat i +3\hat j)\times 2\hat j \times 10^6$$ \Rightarrow \vec F =-2 \times 10^{-6} (2 \hat i\times 2\hat j) \times 10^6$$ \Rightarrow \vec F =-8\hat k N$Therefore, A is correct option.

Question 4

A particle of mass m and charge q has an initial velocity $\vec{v} = v_0 \hat{j}$. If an electric field $\vec{E} = E_0 \hat{i}$ and magnetic field $\vec{B} = B_0 \hat{i}$ act on the particle, its speed will double after a time.

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Given $\vec{V}_i = V_0 \hat{j} , \ \vec{E}  = E_0 \hat{i} , \vec{B} = B_0 \hat{i}$Now we see that the velocity is perpendicular to the magnetic field so the magnitude of velocity will not change due to magnetic field.The magnitude of velocity will change in 'i' direction due to $\vec{E} $ only$acc = \dfrac{force}{mass} $ due to $\vec{E}$ on particle $= \dfrac{q E_0}{m} \hat{i}$Let us say that the speed of particle doubles after time 't'. in y direction the speed will be $V_0$ only in $y - z$ plane always perpendicular to x-direction so net speed at time 't'.$(V_{net})^2 =  \underbrace { (V_0)^2 }_{ V \ in \ y - z \ plane } +  \underbrace { \left(\dfrac{q E_0}{m} t\right)^2 }_{ V_x \ after \ time\ t \, in\ x-direction\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}  $ .Also by given condition $V_{net} = 2V_0$$\Rightarrow (2 V_0)^2 = V_0^2 + \left(\dfrac{q E_0}{m} t\right)^2$$\Rightarrow 4 V_0^2 - V_0^2 = \left(\dfrac{q E_0}{m} t \right)^2$$\Rightarrow 3 V_0^2 = \left(\dfrac{q E_0}{m} t \right)^2$$\Rightarrow t = \dfrac{\sqrt{3} V_0 m}{q E_0}$ Option (A) is correct.

Question 5

A bullet of mass $2g$ is with charge of $2\mu C$. To attain speed of $10ms^{-1}$,it must be accelerated through potential difference of

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Velocity $=\sqrt{\dfrac{2qV}{m}}$$V=\dfrac{mv^2}{2q}$$V=\dfrac{2\times 10^{-3}\times 10\times 10}{2\times 2\times 10^{-6}}$$V=50000V$$V=50kV$

Question 6

An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?

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Hint:- If a charge $q$ is moving with velocity $v$ in a Magnetic Field $B$, such that angle between velocity of charge and Magnetic Field is $\theta$. Then Force acting on charge is given by,$F = qvBsin\theta$Step 1: Note the given values and calculate force on an electron in magnetic field.$\bullet$ It is given that an electron is projected with uniform velocity along the axis of current carrying long solenoid. Let charge on electron is $e$ and uniform velocity of electron is $v$. Also let at any point on axis of solenoid Magnetic Field is $B$.$\bullet$ We know that Magnetic Field on axis of solenoid is always parallel to the axis. And since electron is moving along axis of solenoid, we have $\theta = 0&#176;$ or $\theta = 180&#176;$. In both the cases value of $\sin\theta$ is zero. $\bullet$ Thus net force acting on electron, that is $F = qvBsin\theta = 0$Step 2: Explanation for options:-$\bullet$ Since net force on electron is zero, by Newton's Second Law its acceleration will be zero. Option A is incorrect.$\bullet$ Also since net force on electron is zero, its direction of motion will not change.  So Option B is incorrect.$\bullet$As theta is zero not 45 so option C is incorrect.$\bullet$ As net force is zero so net acceleration is zero. So the electron will continue to move along the axis with same velocity.Thus, only Option D is correct.

Question 7

A charged particle is projected in a magnetic field $\vec{B}=(3\vec{i}+4\vec{j})\times 10^{-2}T$ and the acceleration is found to be $\vec{a}=(x\vec{i}+2\vec{j})m/s^{2}$. The value of x is :

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$\overline{F}=q(\overline V\times\overline B)$So $\overline F\bot\overline B$$\therefore \overline F\cdot \overline B=0$$\overline ma\cdot\overline B=0$$\overline a\cdot\overline B=0$$(x\overline i+2\overline j)\cdot(3\overline i+4\overline j)=0$$3x+8=0$$x=\dfrac{-8}{3}$

Question 8

Write the expression for Lorentz magnetic force on a particle of charge $q$ moving with velocity $v$ in a magnetic field $B$. Shown that two no work is done by this force on the charged particle.

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Lorentz force = magnetic force + electric force F = [ qvb sin$\theta$ + qe ] $\vec F = q(\vec V \times \vec B)$ (1)   $\vec F$  is perpendicular to both $\vec V$ and $\vec B$. if $\vec {ds}$ is the instantaneous displacement of the change $\vec{ds}$ is also perpendicular to $\vec F$$W = \vec F.\vec{ds}$      = Fs $cos90^0$but $cos90^0$ = 0 W = 0 Hence increase in kinetic energy is zero

Question 9

Write the expression for the force on a charge moving in a magnetic field.Use this expression to define the $SI$ unit of magnetic field.

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Force on a charge $(q)$ moving in a magnetic field $B$ with velocity $\overrightarrow { v } $ making an angle $\theta$ (with the direction of magnetic field ($\overrightarrow {  B}$ )is given by$F_m=qvB \sin \theta$When $\theta=90^o \Rightarrow \sin \theta =1$, so  $F_m=qvB$or $B=\dfrac{F_m}{qv}$If $v=1\ m/s, B=\dfrac{F_m}{q}\ newton/coulomb.$SI unit of magnetic field is tesla.Thus, $1$ tesla is the magnetic field in which a charge particle moving with velocity $1\ m/s$ perpendicular to velocity experiences a force of $1 \ newton/coulomb$.

Question 10

Two long and parallel straight wires carrying currents of $2\ A$ and $5$ A in the opposite directions are separated by a distance of $1\ cm$. Find the nature and magnitude of the magnetic force between them.

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$I_1=2 A, I_2=5A, a=1\ cm= 1 \times 10^{-2} \ m$Force between two parallel wire per unit length is given by-$F = \dfrac{\mu _0}{2 \pi}. \dfrac{I_1 I_2}{a} = 2 \times 10^{-7} \times \dfrac{2 \times 5}{1 \times 10^{-2}} = 20 \times 10^{-5}\ N$ (Repulsive).Both will repel because current in both wires is opposite to each other.

Revise with Concepts

Magnetic Field and Magnetic Force

Quick Summary With Stories

Magnetic Force Due To Current Carrying Wire in Magnetic Field

Basics Of Lorentz Force

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

A charged particle moves through a magnetic field perpendicular to its direction. Then

The magnetic force acting on a charged particle of charge $- 2 μ C$ in a magnetic field of 2T acting in y-direction, when the particle velocity is $(2 \overset{ˇ}{i} + 3 \overset{ˇ}{j})$ $\times 1 0^{6} m s^{- 1}$ is :

A particle of mass m and charge q has an initial velocity $v = v_{0} \hat{j}$ . If an electric field $E = E_{0} \hat{i}$ and magnetic field $B = B_{0} \hat{i}$ act on the particle, its speed will double after a time.

A bullet of mass $2 g$ is with charge of $2 μ C$ . To attain speed of $10 m s^{- 1}$ ,it must be accelerated through potential difference of

An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?

A charged particle is projected in a magnetic field $B = (3 i + 4 j) \times 1 0^{- 2} T$ and the acceleration is found to be $a = (x i + 2 j) m / s^{2}$ . The value of x is :

Write the expression for Lorentz magnetic force on a particle of charge $q$ moving with velocity $v$ in a magnetic field $B$ . Shown that two no work is done by this force on the charged particle.

Write the expression for the force on a charge moving in a magnetic field.
Use this expression to define the $S I$ unit of magnetic field.

Two long and parallel straight wires carrying currents of $2 A$ and $5$ A in the opposite directions are separated by a distance of $1 c m$ . Find the nature and magnitude of the magnetic force between them.

Revise with Concepts

Magnetic Field and Magnetic Force

Quick Summary With Stories

Magnetic Force Due To Current Carrying Wire in Magnetic Field

Basics Of Lorentz Force

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

A charged particle moves through a magnetic field perpendicular to its direction. Then

The magnetic force acting on a charged particle of charge−2μC in a magnetic field of 2T acting in y-direction, when the particle velocity is (2iˇ+3jˇ​) ×106ms−1 is :

A particle of mass m and charge q has an initial velocity v=v0​j^​. If an electric field E=E0​i^ and magnetic field B=B0​i^ act on the particle, its speed will double after a time.

A bullet of mass 2g is with charge of 2μC. To attain speed of 10ms−1,it must be accelerated through potential difference of

An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?

A charged particle is projected in a magnetic field B=(3i+4j​)×10−2T and the acceleration is found to be a=(xi+2j​)m/s2. The value of x is :

Write the expression for Lorentz magnetic force on a particle of charge q moving with velocity v in a magnetic field B. Shown that two no work is done by this force on the charged particle.

Write the expression for the force on a charge moving in a magnetic field. Use this expression to define the SI unit of magnetic field.

Two long and parallel straight wires carrying currents of 2 A and 5 A in the opposite directions are separated by a distance of 1 cm. Find the nature and magnitude of the magnetic force between them.

The magnetic force acting on a charged particle of charge $- 2 μ C$ in a magnetic field of 2T acting in y-direction, when the particle velocity is $(2 \overset{ˇ}{i} + 3 \overset{ˇ}{j})$ $\times 1 0^{6} m s^{- 1}$ is :

A particle of mass m and charge q has an initial velocity $v = v_{0} \hat{j}$ . If an electric field $E = E_{0} \hat{i}$ and magnetic field $B = B_{0} \hat{i}$ act on the particle, its speed will double after a time.

A bullet of mass $2 g$ is with charge of $2 μ C$ . To attain speed of $10 m s^{- 1}$ ,it must be accelerated through potential difference of

A charged particle is projected in a magnetic field $B = (3 i + 4 j) \times 1 0^{- 2} T$ and the acceleration is found to be $a = (x i + 2 j) m / s^{2}$ . The value of x is :

Write the expression for Lorentz magnetic force on a particle of charge $q$ moving with velocity $v$ in a magnetic field $B$ . Shown that two no work is done by this force on the charged particle.

Write the expression for the force on a charge moving in a magnetic field.
Use this expression to define the $S I$ unit of magnetic field.

Two long and parallel straight wires carrying currents of $2 A$ and $5$ A in the opposite directions are separated by a distance of $1 c m$ . Find the nature and magnitude of the magnetic force between them.