Q: A proton is fired with a speed of 2\times10^6 m/s at an angle of 60^o to the X- axis. If a uniform magnetic field of 0.1T is applied along the Y- axis, the force acting on the proton is

F=q (\vec {V}\times \vec {B}) \theta=30^o angle between velocity & field =1\cdot 6\times 10^{-19}\times 2\times 10^6\times 0\cdot 1\times \sin 30^o =1\cdot 6\times 10^{-14} N

Q: A particle of charge q and mass m moving with a velocity v along x- axis enters the region x > 0 with uniform magnetic field B along the \hat k direction. The particle will penetrate in this region in the x- direction up to a distance d equal to

Lorentz force F=q(\vec{v}\times\vec{B})=q(v\hat{i}\times B\hat{k}) =-qvB\hat{j} The particle move in a circle, the plane of the circle is perpendicular to the direction of magnetic field. Centripetal force is provided by the Lorentz force. qvB= \dfrac{mv^2}{r} \Rightarrow r =\dfrac{mv}{qB}

Q: A mono energetic electron beam with the speed of 5.2\times 10^{6} ms^{-1} enters into the magnetic field of induction 3\times 10^{-4} T , directed normal to the beam. Then the radius of the circle traced by the beam : (given e/m = 1.76\times 10^{11}C Kg^{-1} )

Magnetic field = 3 \times 10^{-4} Now, Force required for centripetal acceleration =\dfrac{mv^{2}} {r} --------------------- (I)For provided by B towards the center = qvB For the electron to move in a circle, \dfrac{mv^{2}}{r} = qvB [(I) = (II)] \Rightarrow \dfrac{mv}{Bq} = r \Rightarrow r = \dfrac{5.2 \times 10^{6}}{3 \times 10^{-4} \times 1.76 \times 10^{11}} \Rightarrow r = 0.09848 m \Rightarrow r = 0.098m

Question 1

The mono energetic beams of electrons moving along +y direction enter a region of uniform electric and magnetic fields. If the beam goes straight through these simultaneously, then field B and E are directed possibly along

Accepted Answer

\overrightarrow{V} =V({\widehat{j}}) Hence , B and E should be perpendicular to each other and also perpendicular to the velocity . Hence , B and E should be in z - axis and x axis If B is towards + z axis , Force = q(V 	imes B ) is in + x direction . To balance this , E has to be in -x direction

Question 2

A proton is fired with a speed of  2	imes10^6  m/s at an angle of  60^o  to the X- axis. If a uniform magnetic field of  0.1T  is applied along the Y- axis, the force acting on the proton is

Accepted Answer

F=q (\vec {V}	imes \vec {B}) 	heta=30^o  angle between velocity & field =1\cdot 6	imes 10^{-19}	imes 2	imes 10^6	imes 0\cdot 1	imes \sin 30^o =1\cdot 6	imes 10^{-14} N

Question 3

A neutron , a proton, an electron and an  \alpha   particle enter  perpendicular uniform magnetic field, with the same uniform velocity. The path of electron in the following figure will be :

Accepted Answer

The neutron goes undeflected, the  \alpha   and proton being  both positively charged moves towards left and electron being negatively charged moves towards right. or we can also use right hand to find  the direction of force using formula   F = q(V	imes B)  but it moves right as  q  is  -ve.

Question 4

A  \beta -  particle enters a magnetic field making an angle of  45^{o}  with the field lines. The path of the particle is

Accepted Answer

the  v_x = \dfrac{v}{\sqrt 2}   will be responsible for circular motion and  v_y  will take it in  +Y  direction thus making it a spiral path

Question 5

A particle of charge  q  and mass  m  moving with a velocity  v  along  x- axis enters the region  x > 0  with uniform magnetic field  B  along the  \hat k  direction. The particle will penetrate in this region in the  x- direction up to a distance  d  equal to

Accepted Answer

Lorentz force  F=q(\vec{v}	imes\vec{B})=q(v\hat{i}	imes B\hat{k}) =-qvB\hat{j} The particle move in a circle, the plane of the circle is perpendicular to the direction of magnetic field. Centripetal force is provided  by the  Lorentz force. qvB= \dfrac{mv^2}{r} \Rightarrow  r =\dfrac{mv}{qB}

Question 6

A mono energetic electron beam with the speed of  5.2	imes 10^{6} ms^{-1}  enters into the magnetic field of induction  3	imes 10^{-4} T , directed normal to the beam. Then the radius of the circle traced by the beam : (given  e/m = 1.76	imes 10^{11}C Kg^{-1} )

Accepted Answer

Magnetic field  = 3 	imes 10^{-4} Now,  Force required for centripetal acceleration  =\dfrac{mv^{2}} {r}  --------------------- (I)For provided by B towards the center  = qvB For the electron to move in a circle,  \dfrac{mv^{2}}{r} = qvB                   [(I) = (II)] \Rightarrow \dfrac{mv}{Bq} = r \Rightarrow r = \dfrac{5.2 	imes 10^{6}}{3 	imes 10^{-4} 	imes 1.76 	imes 10^{11}} \Rightarrow r = 0.09848 m \Rightarrow r = 0.098m

Question 7

The resistance of a galvanometer is  50\Omega   and the current required to give full-scale deflection is  100\mu A . In order to convert it into an ammeter, reading upto  10A , it is necessary to put resistance of:

Accepted Answer

For a galvanometer to be converted into ammeter it should be connected in parellel thus, Resistance is parallel  \quad S=\cfrac { G{ I }_{ g } }{ I-{ I }_{ g } } =\cfrac { 50	imes 100	imes { 10 }^{ -6 } }{ \left( 10-100	imes { 10 }^{ -6 } ight)  }  \Rightarrow S=5	imes { 10 }^{ -4 }\Omega

Question 8

A galvanometer coil has a resistance of  50 \Omega  and the meter shows full scale deflection for a current of  5 mA . This galvanometer is converted into voltmeter of range  0 - 20 V  by connecting

Accepted Answer

Given :    R_G = 50\Omega                 I_g = 5mA = 0.005 A            V = 20  voltsA galvanometer is converted into a voltmeter by connecting a resistor in series with resistance of galvanometer.Let the resistor to be connected in series with  R_G  be  R . 	herefore      V = I_g (R_G+R) Or      20 = 0.005(50+R) \implies     R = 4000 - 50 = 3950\Omega

Question 9

Two long parallel conductors are placed at right angles to a metre scale at the  2 cm  and  4 cm  marks, as shown in the figure. They carry currents of  1 A  and  3 A  respectively. They will produce zero magnetic field at the (ignore the Earth's magnetic field)

Accepted Answer

|B|_{net}=0 (\dfrac{\mu _{0}i_{1}}{2\pi r_{1}})-(\dfrac{\mu _{0}i_{2}}{2\pi r_{2}})=0            where  r1  and  r_2  are distances from wires respectively. Assuming the distance from wire 1 is  x , then distance from wire 2 will be  2-x \dfrac{i_{1}}{x}=\dfrac{i_2}{2-x} \dfrac{1}{x}=\dfrac{3}{2-x} 4x=2 x=0.5 cm  from  1A  wire.distance from the y-axis  2+0.5=2.5cm  .

Question 10

Three long straight wires A, B and C are carrying currents as shown in figure. Then the resultant force on B is directed :

Accepted Answer

Hint:If the current in the two parallel wires is in the same direction then the wires attract each other.Step 1: Expression of force between two parallel wires,The force between two parallel wires having current  I_1  and  I_2  and separated by distance  d  is, F=\frac{\mu_oI_1 I_2}{2\pi d} Therefore, F\propto I_1I_2  and  F\propto \frac{1}{d} Step 2: Find the stronger force.Wire A and wire C are trying to pull wire B towards them.The separation between the wires for both cases is the same. I_A I_B=(1A)(2A)=3A^2 I_c I_B=(3A)(2A)=6A^2 \because   F\propto I_1I_2 	herefore   F_{BC}>F_{AB} Therefore wire is pulled towards  C . 	extbf{Hence option D correct}

Question 11

A current of i amp flows in a loop having circular arc of radius r subtending an angle  	heta   as shown in the figure. The magnetic field at the centre of the circle is:

Accepted Answer

The wire AOB will not contribute in magnetic field at 0 as it is along the length of the wire.For the rest of the wire,Magnetic field at centre if  2\pi  angle is covered   B = \dfrac{\mu_oi}{2r} Magnetic field at the centre if  	heta  radian is covered  B = \dfrac{\mu_oi	imes 	heta}{2r	imes2\pi} = \dfrac{\mu_o i 	heta}{4 \pi r}

Question 12

Two circular coils  X  and  Y  have equal number of turns and carry equal currents in the same sense and subtend same solid angle at point  (O) .  If the smaller coil  X  is midway between   O  and  Y,   then if we represent the magnetic induction due to bigger coil  Y   at  O  as  B_y   and that due to smaller coil  X  at  O  as  B_x , then

Accepted Answer

We know magnetic induction due to a circular currrent carrying coil on its axis at distance x from its center is given by, \displaystyle B=\dfrac { { \mu  }_{ o }nI{ R }^{ 2 } }{ 2{ ({ x }^{ 2 }+{ R }^{ 2 }) }^{ 3/2 } }  Now Let larger coil be at a distance  x  and have radius  R . then smaller coil will be have radius  R/2  and at a distance  x/2  since it is midway and subtends same angle as subtended by larger coil at point O.Magnetic field due to larger coil at O B_y=\dfrac { { \mu  }_{ o }nI{ R }^{ 2 } }{ 2{ ({ x }^{ 2 }+{ R }^{ 2 }) }^{ 3/2 } }  and due to smaller coil will be B_x=\dfrac { { \mu  }_{ o }nI{ (R/2) }^{ 2 } }{ 2{ ({ (x/2) }^{ 2 }+{ (R/2) }^{ 2 }) }^{ 3/2 } }  B_{ x }=\dfrac { { \mu  }_{ o }nI{ (R) }^{ 2 } }{ 4{ ({ (x) }^{ 2 }+{ (R) }^{ 2 }) }^{ 3/2 } }  Therefore,  B_y/B_x=1/2 option(C)

Question 13

Two long straight wires are connected by a circular section which has a radius  R  .All the three segments lie in the same plane and carry a current  I . The magnetic induction at the centre  O  of the circular segments is :

Accepted Answer

magnetic field due to the wire =0 And CD & BA are passing through  O So due to straight wires  B=0 B  due to circular arc = \dfrac{\mu_0 i}{4\pi r} (	heta) =\dfrac{\mu_0i \alpha}{4\pi r}

Question 14

The torque acting on the loop as shown in the figure is :

Accepted Answer

Torque on the loop is, 	au =NBAI\sin	heta =1	imes10	imes0.4	imes50	imes10^{-4}	imes \sin90^\circ{} =2	imes10^{-2} N-m   along  -\ x  axis

Question 15

A circular loop of radius  20   cm  is placed in a uniform magnetic field  \vec{B}=2 T  in X-Y plane. The loop carries a current  1   A  in the direction shown in the figure. The magnitude of torque acting on the loop is nearly :

Accepted Answer

Torque on loop is, 	au=NIBA \sin	heta                        (	heta  is angle between magnetic field  B  and Area Vector) =1	imes1	imes2	imes A	imes0.04	imes \sin90^o =0.25 Nm

Moving Charges And Magnetism