Q: A current of i ampere is flowing through each ofthe bent wires as shown the magnitude and direction of magnetic field at O is

\$B=\dfrac { { \mu }_{ 0 }I }{ 2R } \dfrac { 3 }{ 4 } +\dfrac { { \mu }_{ 0 }I }{ 2R } \frac { 1 }{ 4 } \$\$B=\dfrac { { \mu }_{ 0 }I }{ 8 } \left[ \dfrac { 3 }{ { R }^{ \prime } } +\dfrac { 1 }{ R } \right] \$

Q: \$\mathrm{A}\$ magnetic field \$\vec{\mathrm{B}}=\mathrm{B}_{0}\hat{\mathrm{j}}\$ exists in the region a \$<\mathrm{x}<2\mathrm{a}\$ and \$\vec{\mathrm{B}}=-\mathrm{B}_{0}\hat{\mathrm{j}}\$ , in the region \$2\mathrm{a}<\mathrm{x}<3\mathrm{a}\$, where \$\mathrm{B}_{0}\$ is a positive constant. \$\mathrm{A}\$ positive point charge moving with a velocity \$\vec{\mathrm{v}}=\mathrm{v}_{0}\hat{\mathrm{i}}\$, where \$\mathrm{v}_{0}\$ is a positive constant, enters the magnetic field at \$\mathrm{x}=\mathrm{a}\$. The trajectory of the charge in this region can be like,

\$F=q vB\$Now \$\bar v = v_0 \hat i , \bar B = B_0 \hat j Thus, \bar F = v_0B_0 \hat k\$Thus the particle will start straying towards +z direction.After it reaches x=2a, \$\bar F = - v_0B_0 \hat k\$There won't be any jerks as force always acts perpendicular to particle motion, and acceleration is finite. Now it starts to stray towards -z direction.

Q: An electron is projected with velocity \$\mathrm{v}_{0}\$ in a uniform electric field \$\mathrm{E}\$ perpendicular to the field. Again it is projected with velocity \$\mathrm{v}_{0}\$ perpendicular to a uniform magnetic field \$\mathrm{B}\$. If \$r_1\$ is initial radius of curvature just after entering in the electric field and \$\mathrm{r}_{2}\$ is initial radius of curvature just after entering in magnetic field then the ratio \$\mathrm{r}_{1}/\mathrm{r}_{2}\$ is equal to :

For electric field,Refer image 1.\$q\rightarrow\$ change of electron.Four due to electric field \$=qE\$\$\therefore \dfrac{mv_{0^2}}{R_e}=qE\$where \$R_e\$ is radius of curvature,\$\therefore \dfrac{1}{R_e}=\dfrac{qE}{mv_{0^2}}\$\$\therefore =\dfrac{mv_{0^2}}{qE}=r_1\$In magnetic field,Focus on electron \$=qV_0B\$Refer image 2.Balancing of centripetal and centrifugal focuses,\$\dfrac{mV_{0^2}}{Rb}=qV_0B\$ \$\therefore R_b=\dfrac{mV_0}{qB}=r_2\$\$\therefore \dfrac{r_1}{r_2}=\dfrac{R_c}{R_b}=\dfrac{\dfrac{mV_{0^2}}{qE}}{\dfrac{mV_0}{qB}}\$\$\therefore \dfrac{r_1}{r_2}=\dfrac{BV_0}{E}\$Answer is option \$D\$

Q: A charged particle q is moving with a velocity \$\vec{v_{1}}=2\hat{i}m/s\$ at a point in a magnetic field \$\vec{B}\$ and experiences a force \$\vec{F_{1}}=q(\hat{k}-2\hat{j})N.\$ If the same charge moves with velocity \$\vec{v_{2}}=2\hat{j}m/s\$ from the same point in that magnetic field and experiences a force \$\vec{F_{2}}=q(2\hat{i}+\hat{k})N\$, the magnetic induction at that point will be :

Let the magnetic induction at required point be \$\bar { B } ={ B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \$Magnetic force \$\left( \bar { F } \right) \$ on particle of charge (q), moving with velocity \$\bar { V } \$ in the magnetic field region of intensity \$\left( \bar { B } \right) \$ is given as\$\bar { F } =q\left( \bar { V } \times \bar { B } \right) \$Case 1: \$\bar { { F }_{ 1 } } =q\left( \bar { { V }_{ 1 } } \times \bar { B } \right) \$\$q\left( \hat { k } -2\hat { j } \right) =q\left( 2\hat { i } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right) \$\$\hat { k } -2\hat { j } =2{ B }_{ y }\hat { k } -2{ B }_{ z }\hat { j } \$\$\therefore 2{ B }_{ y }=1,\quad { B }_{ y }={ 1 }/{ 2 }.....\left( i \right) \$\$-2=-2{ B }_{ z },\therefore { B }_{ z }=1.....\left( ii \right) \$Case 2: \$\bar { { F }_{ 2 } } =q\left( \bar { { V }_{ 2 } } \times \bar { B } \right) \$\$q\left( 2\hat { i } +\hat { k } \right) =q\left( 2\hat { j } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right) \$\$2\hat { i } +\hat { k } =-2{ B }_{ x }\hat { k } +2{ B }_{ z }\hat { i } \$\$\therefore -2{ B }_{ x }=1,\quad \therefore +{ B }_{ x }={ 1 }/{ 2 }......\left( iii \right) \$Thus from \$\left( i \right) ,\left( ii \right) \& \left( iii \right) \$\$\bar { B } =-\cfrac { 1 }{ 2 } \hat { i } +\cfrac { 1 }{ 2 } \hat { j } +\hat { k } \$

Q: A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance

Given \$A\$ and \$B\$ are two like parallel forces. A couple of moment \$H\$ liesin the Plane of \$A\$ and \$B\$ and is contained with them.\$\left( A+B \right) d=H\$So the resultant of \$A\$ and \$B\$after combining is displaced through a distance Hence the correct answer is \$d=\$ \$\cfrac { H }{ \left( A+B \right) } \$

Q: A particle of charge \$-16\times 10^{-18}\$ coulomb moving with velocity 10 \$\mathrm{m}\mathrm{s}^{-1}\$ along the x-axis enters a region where a magnetic field of induction \$\mathrm{B}\$ is along the \$\mathrm{y}\$ -axis, and an electric field of induction \$\mathrm{B}\$ is along the \$\mathrm{y}\$-axis, and an electric field of magnitude \$10^{4}\mathrm{V}/\mathrm{m}\$ is along the negative \$\mathrm{z}\$-axis. lf the charged particle continues moving along the \$\mathrm{x}\$-axis, the magnitude of \$\mathrm{B}\$ is

Particle travels along \$x\$ axis Hence \${ V }_{ y }={ V }_{ z }=0\$Electric feild is along the negative \$z\$ axis\${ E }_{ x }={ E }_{ y }=0\$Therefore net force on the particle \$\vec { F } =q\left( \vec { E } +V\times \vec { B } \right) \$Resolve the motion along three coordinate axis\${ a }_{ x }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ x }+{ V }_{ y }{ B }_{ z }-{ V }_{ x }{ B }_{ y } \right) \\ { a }_{ y }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ y }+{ V }_{ z }{ B }_{ x }-{ V }_{ xz } \right) \\ { a }_{ z }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ z }+{ V }_{ x }{ B }_{ y }-{ V }_{ y }{ B }_{ x } \right) \$Since, \${ V }_{ y }={ V }_{ z }=0\\ { E }_{ x }={ E }_{ y }=0\\ { B }_{ x }={ B }_{ z }=0\$\${ a }_{ x }={ a }_{ y }=0\\ { a }_{ z }=\cfrac { q }{ m } \left( -{ E }_{ z }+{ V }_{ x }{ B }_{ y } \right) \$Again \${ a }_{ z }=0\$ , as the particle transverse through the region undeflected \${ E }_{ z }={ V }_{ x }{ B }_{ y }\$\${ B }_{ y }=\cfrac { { E }_{ z } }{ { V }_{ x } } =\cfrac { { 10 }^{ 4 } }{ 10 } ={ 10 }^{ 3 }wb/m^{ 2 }\$

Q: A charged particle having charge q experiences a force \$\vec{F}=q(-\vec{j}+\vec{k})N\$ in a magnetic field B when it has a velocity \$v_{1} = 1\hat i \ m/s\$. The force becomes \$\vec{F}=q(\vec{i}-\vec{k})N\$ when the velocity is changed to \$v_{2}=1\hat{j}m/sec\$. The magnetic induction vector at that point is :

Let the magnetic induction at required point be \$\bar { B } =\left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) T\$Magnetic Force \$\left( \bar { F } \right) \$ on charged particle (q) moving with Velocity (V) in magnetic field \$\left( \bar { B } \right) \$ is given as \$\bar { F } =q\left( \bar { V } \times \bar { B } \right) \$Case 1: \$q\left( -\hat { j } +\hat { k } \right) =q\left( \hat { i } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right) \$\$-\hat { j } +\hat { k } ={ B }_{ y }\hat { k } -{ B }_{ z }\hat { j } \$\${ B }_{ z }=1....\left( i \right) \$\${ B }_{ y }=+1....\left( ii \right) \$Case 2: \$q\left( \hat { j } -\hat { k } \right) =q\left( \hat { j } \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right) \$\$\therefore \hat { j } -\hat { k } =-{ B }_{ x }\hat { k } +{ B }_{ z }\hat { i } \$\$\therefore { B }_{ x }=1....\left( iii \right) \$From \$\left( i \right) ,\left( ii \right) \& \left( iii \right) \$\$\bar { B } =\left( \hat { i } +\hat { j } +\hat { k } \right) T\$ Thus, option A

Q: A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.

\$\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { q\overrightarrow { v } \times \overrightarrow { r } }{ { r }^{ 3 } } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { qvsin\theta }{ { r }^{ 2 } } \$ =\$\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { 2\times 100\times sin{ 30 }^{ 0 } }{ 4 } \$ =\${ 10 }^{ -7 }\times 25\$

Question 1

A current of i ampere is flowing through each ofthe bent wires as shown the magnitude and direction of magnetic field at O is

Accepted Answer

$B=\dfrac { { \mu  }_{ 0 }I }{ 2R } \dfrac { 3 }{ 4 } +\dfrac { { \mu  }_{ 0 }I }{ 2R } \frac { 1 }{ 4 } $$B=\dfrac { { \mu  }_{ 0 }I }{ 8 } \left[ \dfrac { 3 }{ { R }^{ \prime  } } +\dfrac { 1 }{ R }  \right] $

Question 2

$\mathrm{A}$ magnetic field $\vec{\mathrm{B}}=\mathrm{B}_{0}\hat{\mathrm{j}}$ exists in the region a $<\mathrm{x}<2\mathrm{a}$ and $\vec{\mathrm{B}}=-\mathrm{B}_{0}\hat{\mathrm{j}}$ , in the region $2\mathrm{a}<\mathrm{x}<3\mathrm{a}$, where $\mathrm{B}_{0}$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\vec{\mathrm{v}}=\mathrm{v}_{0}\hat{\mathrm{i}}$, where $\mathrm{v}_{0}$ is a positive constant, enters the magnetic field at $\mathrm{x}=\mathrm{a}$. The trajectory of the charge in this region can be like,

Accepted Answer

$F=q vB$Now $\bar v = v_0 \hat i , \bar B = B_0 \hat j Thus, \bar F = v_0B_0 \hat k$Thus the particle will start straying towards +z direction.After it reaches x=2a, $\bar F = - v_0B_0 \hat k$There won't be any jerks as force always acts perpendicular to particle motion, and acceleration is finite. Now it starts to stray towards -z direction.

Question 3

An electron is projected with velocity $v_{0}$ in a uniform electric field $E$ perpendicular to the field. Again it is projected with velocity $v_{0}$ perpendicular to a uniform magnetic field $B$ . If $r_{1}$ is initial radius of curvature just after entering in the electric field and $r_{2}$ is initial radius of curvature just after entering in magnetic field then the ratio $r_{1} / r_{2}$ is equal to :

A

\frac{B v _{0}^{2}}{E}

B

\frac{B}{E}

C

\frac{E v _{0}}{B}

D

\frac{B v _{0}}{E}

Hard

View solution>

Accepted Answer

For electric field,Refer image 1.$q\rightarrow$ change of electron.Four due to electric field $=qE$$\therefore \dfrac{mv_{0^2}}{R_e}=qE$where $R_e$ is radius of curvature,$\therefore \dfrac{1}{R_e}=\dfrac{qE}{mv_{0^2}}$$\therefore =\dfrac{mv_{0^2}}{qE}=r_1$In magnetic field,Focus on electron $=qV_0B$Refer image 2.Balancing of centripetal and centrifugal focuses,$\dfrac{mV_{0^2}}{Rb}=qV_0B$ $\therefore R_b=\dfrac{mV_0}{qB}=r_2$$\therefore \dfrac{r_1}{r_2}=\dfrac{R_c}{R_b}=\dfrac{\dfrac{mV_{0^2}}{qE}}{\dfrac{mV_0}{qB}}$$\therefore \dfrac{r_1}{r_2}=\dfrac{BV_0}{E}$Answer is option $D$

Question 4

A charged particle q is moving with a velocity $v_{1} = 2 \hat{i} m / s$ at a point in a magnetic field $B$ and experiences a force $F_{1} = q (\hat{k} - 2 \hat{j}) N .$ If the same charge moves with velocity $v_{2} = 2 \hat{j} m / s$ from the same point in that magnetic field and experiences a force $F_{2} = q (2 \hat{i} + \hat{k}) N$ , the magnetic induction at that point will be :

Accepted Answer

Let the magnetic induction at required point be $\bar { B } ={ B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } $Magnetic force $\left( \bar { F }  \right) $ on particle of charge (q), moving with velocity $\bar { V } $ in the magnetic field region of intensity $\left( \bar { B }  \right) $ is given as$\bar { F } =q\left( \bar { V } \times \bar { B }  \right) $Case 1: $\bar { { F }_{ 1 } } =q\left( \bar { { V }_{ 1 } } \times \bar { B }  \right) $$q\left( \hat { k } -2\hat { j }  \right) =q\left( 2\hat { i } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k }  \right)  \right) $$\hat { k } -2\hat { j } =2{ B }_{ y }\hat { k } -2{ B }_{ z }\hat { j } $$\therefore 2{ B }_{ y }=1,\quad { B }_{ y }={ 1 }/{ 2 }.....\left( i \right) $$-2=-2{ B }_{ z },\therefore { B }_{ z }=1.....\left( ii \right) $Case 2: $\bar { { F }_{ 2 } } =q\left( \bar { { V }_{ 2 } } \times \bar { B }  \right) $$q\left( 2\hat { i } +\hat { k }  \right) =q\left( 2\hat { j } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k }  \right)  \right) $$2\hat { i } +\hat { k } =-2{ B }_{ x }\hat { k } +2{ B }_{ z }\hat { i } $$\therefore -2{ B }_{ x }=1,\quad \therefore +{ B }_{ x }={ 1 }/{ 2 }......\left( iii \right) $Thus from $\left( i \right) ,\left( ii \right) \& \left( iii \right) $$\bar { B } =-\cfrac { 1 }{ 2 } \hat { i } +\cfrac { 1 }{ 2 } \hat { j } +\hat { k } $

Question 5

A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance

Accepted Answer

Given $A$ and $B$ are two like parallel forces. A couple of moment $H$ liesin the Plane of $A$ and $B$ and is contained with them.$\left( A+B \right) d=H$So the resultant of $A$ and $B$after combining is displaced through a distance Hence the correct answer is $d=$ $\cfrac { H }{ \left( A+B \right)  } $

Question 6

The magnitude of force per unit length on a wire carrying current $I$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $L$ and current in them is $I$ ) is:

Accepted Answer

Net Magnetic field at point O due to both the wires shall be$\displaystyle B=-2\int_{90^{0}}^{0^{0}}\frac{\mu_0 I}{2\pi l} \sin\theta d\theta$   along $-k$$\displaystyle \therefore B= -\dfrac{\mu_0 I}{\pi l}$ along $-k$ Now, $F=I(\vec{dl} \times \vec{B})$Where, $\vec{dl}$ is along $i$Therefore, $\dfrac{F}{dl}= \dfrac{\mu I^{2}}{\pi l}$  along $j$

Question 7

A particle of charge $-16\times 10^{-18}$ coulomb moving with velocity 10 $\mathrm{m}\mathrm{s}^{-1}$ along the x-axis enters a region where a magnetic field of induction $\mathrm{B}$ is along the $\mathrm{y}$ -axis, and an electric field of induction $\mathrm{B}$ is along the $\mathrm{y}$-axis, and an electric field of magnitude $10^{4}\mathrm{V}/\mathrm{m}$ is along the negative $\mathrm{z}$-axis. lf the charged particle continues moving along the $\mathrm{x}$-axis, the magnitude of $\mathrm{B}$ is

Accepted Answer

Particle travels along $x$ axis Hence ${ V }_{ y }={ V }_{ z }=0$Electric feild is along the negative $z$ axis${ E }_{ x }={ E }_{ y }=0$Therefore net force on the particle $\vec { F } =q\left( \vec { E } +V\times \vec { B }  \right) $Resolve the motion along three coordinate axis${ a }_{ x }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ x }+{ V }_{ y }{ B }_{ z }-{ V }_{ x }{ B }_{ y } \right) \ { a }_{ y }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ y }+{ V }_{ z }{ B }_{ x }-{ V }_{ xz } \right) \ { a }_{ z }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ z }+{ V }_{ x }{ B }_{ y }-{ V }_{ y }{ B }_{ x } \right) $Since, ${ V }_{ y }={ V }_{ z }=0\ { E }_{ x }={ E }_{ y }=0\ { B }_{ x }={ B }_{ z }=0$${ a }_{ x }={ a }_{ y }=0\ { a }_{ z }=\cfrac { q }{ m } \left( -{ E }_{ z }+{ V }_{ x }{ B }_{ y } \right) $Again ${ a }_{ z }=0$ , as the particle transverse through the region undeflected ${ E }_{ z }={ V }_{ x }{ B }_{ y }$${ B }_{ y }=\cfrac { { E }_{ z } }{ { V }_{ x } } =\cfrac { { 10 }^{ 4 } }{ 10 } ={ 10 }^{ 3 }wb/m^{ 2 }$

Question 8

Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are $\vec{v}=3\hat{i}+4\hat{j}$ and $\vec{a}=2\hat{i}+x\hat{j}$. Select the correct alternative(s) :

Accepted Answer

Force & velocity are perpendicular to each other in magnetic fieldSo$\vec{V}\cdot\vec{a}=0$$(3\hat{i}+4\hat{j})\cdot(2\hat{i}+x\hat{j})=0$$6+4x=0$$x=-1.5$No work is done by magnetic field. So, KE of particle is constant

Question 9

A charged particle having charge q experiences a force $F = q (- j + k) N$ in a magnetic field B when it has a velocity $v_{1} = 1 \hat{i} m / s$ . The force becomes $F = q (i - k) N$ when the velocity is changed to $v_{2} = 1 \hat{j} m / s e c$ . The magnetic induction vector at that point is :

Accepted Answer

Let the magnetic induction at required point be $\bar { B } =\left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k }  \right) T$Magnetic Force $\left( \bar { F }  \right) $ on charged particle (q) moving with Velocity (V) in magnetic field $\left( \bar { B }  \right) $ is given as $\bar { F } =q\left( \bar { V } \times \bar { B }  \right) $Case 1: $q\left( -\hat { j } +\hat { k }  \right) =q\left( \hat { i } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k }  \right)  \right) $$-\hat { j } +\hat { k } ={ B }_{ y }\hat { k } -{ B }_{ z }\hat { j } $${ B }_{ z }=1....\left( i \right) $${ B }_{ y }=+1....\left( ii \right) $Case 2: $q\left( \hat { j } -\hat { k }  \right) =q\left( \hat { j } \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k }  \right)  \right) $$\therefore \hat { j } -\hat { k } =-{ B }_{ x }\hat { k } +{ B }_{ z }\hat { i } $$\therefore { B }_{ x }=1....\left( iii \right) $From $\left( i \right) ,\left( ii \right) \& \left( iii \right) $$\bar { B } =\left( \hat { i } +\hat { j } +\hat { k }  \right) T$ Thus, option A

Question 10

A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.

Accepted Answer

$\dfrac { { \mu  }_{ 0 } }{ 4\pi  } \dfrac { q\overrightarrow { v } \times \overrightarrow { r }  }{ { r }^{ 3 } } =\dfrac { { \mu  }_{ 0 } }{ 4\pi  } \dfrac { qvsin\theta  }{ { r }^{ 2 } } $                       =$\dfrac { { \mu  }_{ 0 } }{ 4\pi  } \dfrac { 2\times 100\times sin{ 30 }^{ 0 } }{ 4 } $                       =${ 10 }^{ -7 }\times 25$

Moving Charges And Magnetism

Physics
3838 Views

A current of i ampere is flowing through each ofthe bent wires as shown the magnitude and direction of magnetic field at O is

A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance

The magnitude of force per unit length on a wire carrying current $I$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $L$ and current in them is $I$ ) is:

Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are $v = 3 \hat{i} + 4 \hat{j}$ and $a = 2 \hat{i} + x \hat{j}$ . Select the correct alternative(s) :

A charged particle having charge q experiences a force $F = q (- j + k) N$ in a magnetic field B when it has a velocity $v_{1} = 1 \hat{i} m / s$ . The force becomes $F = q (i - k) N$ when the velocity is changed to $v_{2} = 1 \hat{j} m / s e c$ . The magnetic induction vector at that point is :

A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.

Moving Charges And Magnetism

Physics 3838 Views

A current of i ampere is flowing through each ofthe bent wires as shown the magnitude and direction of magnetic field at O is

A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance

The magnitude of force per unit length on a wire carrying current I at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to L and current in them is I) is:

Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are v=3i^+4j^​ and a=2i^+xj^​. Select the correct alternative(s) :

A charged particle having charge q experiences a force F=q(−j​+k)N in a magnetic field B when it has a velocity v1​=1i^ m/s. The force becomes F=q(i−k)N when the velocity is changed to v2​=1j^​m/sec. The magnetic induction vector at that point is :

A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.

Physics
3838 Views

The magnitude of force per unit length on a wire carrying current $I$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $L$ and current in them is $I$ ) is:

Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are $v = 3 \hat{i} + 4 \hat{j}$ and $a = 2 \hat{i} + x \hat{j}$ . Select the correct alternative(s) :

A charged particle having charge q experiences a force $F = q (- j + k) N$ in a magnetic field B when it has a velocity $v_{1} = 1 \hat{i} m / s$ . The force becomes $F = q (i - k) N$ when the velocity is changed to $v_{2} = 1 \hat{j} m / s e c$ . The magnetic induction vector at that point is :