Multiple Choice questions on Magnetic field and magnetic effects of current for JEE Main and Advanced
Question 1
Find the force on the straight conductor of length .30 m carrying a current of 5 .0 A in the – k direction where the magnetic Field is given by B=3.5 ×10-3 (i - j) T Solution
Force is given by
$\mathbf{F}= I (\mathbf{L} \times \mathbf{B})$
$= (5.0) [(.3)(-\mathbf{k}) \times (3.50 \times 10^{-3})( \mathbf{i}-\mathbf{j}) ]$
$= 7.42 \times 10^{-3}\left ( \frac{-\mathbf{i}-\mathbf{j}}{\sqrt 2} \right )$
Question 2
A current sheet having current K A/m is placed at in the x-y plane at z=0.The direction of current is -j
Find the Magnetic Field at on the z axis Solution
The biot-savart law and consideration of symmetry shows that magnetic field at any point on the z-axis must has only x-components.
This can be proved as
Let’s take point T on the z-axis. Consider two strips R and S of the sheets situated symmetrically on the two side of the point T on the z-axis. Now from biot savart law we know the magnetic field at T due to strip R will be be perpendicular to the line joining the strip R and T. Similarly, we know the magnetic field at T due to strip S will be be perpendicular to the line joining the strip S and T. The resultant of these two will be parallel to the x-axis
Now let’s take a contour 12341 as shown in the figure and let B be the magnitude of the Magnetic field
From Amperes law, we know that
$\oint \mathbf{B}.d\mathbf{l}=\mu _0i$
Applying this law in the closed contour 12341
B(2a) +0+B(2a) + 0=μ0 K2a
Or
$B= \frac {1}{2} \mu _0 K$
The direction of the field is –i
So
Question 3
A current sheet where current per unit length is 6.0 A/m and direction is towards +i is placed at z=0 in the x-y plane. A current wire is located at y=0 and z=5 m. It is placed along the x axis. Find the current and its direction if Magnetic field is zero at the point P whose location is (0, 0, 2.5) m
(a) 47 A, i
(b) 47.1 A, -i
(c) 50 A, i
(d) 50 A, -i Solution
From last question, we know that Magnetic field is independent of the distance for the current sheet.
So, Magnetic field at point P
$B_1=\frac {1}{2} \mu _{0}K$
$=3\mu _{0}$
Direction will be –j. This can be find by applying the right-hand rule to the few current elements in the sheet
Now Magnetic field due to current wire is given by
$B_2=\frac {\mu _0 I }{2\pi d}$
Or
$B_2=\frac {\mu _0 I }{2\pi 2.5}$
So B1=B2
Or I=47.1A
The direction of the current is the wire should be towards i to cancel the magnetic field of the current sheet
Questions 4
A charged particle whose mass is 19.9 ×10-27 kg and charge is 1.6×10-19 C moves with a speed of 3×105 m/s at right angle to a magnetic field of .75 T.
What is the force acting on the charge, centripetal acceleration and radius of the circle in which charged particle moves
(a) 3.6×10-14 N,1.81×1012 m/s2, 49.7×10-3 m
(b) 3.0×10-14 N,1.81×1012 m/s2, 49.7×10-3 m
(c) 3.0×10-14 N,1.8×1012 m/s2, 49.7×10-3 mm
(d) None of the above Solution
Given is
B=.75 T, v=3×105 m/s q=1.6×10-19 C m=19.9×10-27 kg
Force is given by
F=qvB
Substituting the values
F=3.6×10-14 N
Centripetal acceleration is given by
a=F/m
Substituting the values
a=1.81×1012 m/s2
Now we know that
a=v2/r
or r=v2/a
Substituting the values
R=49.7 ×10-3 m
Question 5
Which of the following is true?
(a) Magnetic field exerts force only a moving charge.
(b) Electric field exerts force on both stationary and moving charge
(c) Magnetic field exerts force on charge moving parallel to the direction of the field
(d) All the above Solution
A and B
Question 6
Four Particles P1(proton), P2(electron), P3(alpha), P4(neutron) enters a region of constant magnetic field with same velocities. The magnetic field is perpendicular to the velocity
Match the following Column A
(a) what all particle will not be deflected by the Magnetic field
(b) What all particles will be deflected by the Magnetic Field?
(c) Which particle move in a circular path of maximum radius
(d) Which particle will experience maximum Force? Column B
(p) P1
(q) P2
(r) P3
(s) P4 Solution
Since Neutron does not carry any charge, it won’t get any deflection.
All other three particles carry charge so they will get the deflection
Now radius is given by
R=mv/Bq
For v/B is constant so r depends on the ration m/q
Now we know that
So, r will be maximum for alpha particle
Now force is given by
F=qvB
Now
So, alpha particle will have experienced the maximum force
Question 7
The force experienced by a particle of charge q moving with a velocity v in a uniform magnetic field B is given by F=q(vXB)
Statement I: The vectors F and v are perpendicular to each other
Statement II: The vectors v and B can have any angle between them
Statement III: The vectors F and B are perpendicular to each other
(a) All the statements are true
(b) Statement I and II are true only
(c) Statement I and III are true only
(d) Statement II and III are true only Solution
Since F is derived by the cross product on v and B. So, it will always be Perpendicular to each other.
Question 8
A long wire is bent as shown in figure. Find the magnitude and direction of the magnetic field at the center of the circular part if a current of I amp is passed through the wire
The different parts of the wire do not touch each other at Q Solution
The magnitude of the magnetic field at center due to the straight line is given by
$B_1=\frac {\mu _0 I }{2\pi a}$
B1 is perpendicular to the plane of the page, directed upwards
The field at the center due to the current loop of radius a is
$B_2=\frac {\mu _0 I }{2 a}$
B2 is also perpendicular to the page, directly upward
Therefore, Resultant Magnetic field at the center
=B1+B2
$=\frac {\mu _0 I }{2 a}(\frac {1}{\pi} +1)$
Direction is directed upwards to the plane of the pages
Question 9
A non-conducting thin disc of Diameter D charged uniformly over one side with surface density$\sigma $ rotates about its axis with an angular velocity $\omega $
What is the magnetic field at the center of the disc? Solution
Let us take a ring element of radius r and thickness dr.
The charge on the ring element
$dq=\sigma 2 \pi rdr$
Since the disc is rotating, charge is also rotating.
So, current due to the element
$di=\frac {(\sigma 2 \pi rdr) \omega }{2\pi }=\sigma \omega rdr $
Now we know that magnetic field at the center due to circular current ring is
$dB=\frac {\mu _0 di}{2r}$
The disk can have assumed to be made of these infinite elements
So, Magnetic Field at the center can be given as
$B=\int dB=\int_{0}^{d/2} \frac {\mu _0 \sigma \omega rdr}{2r} = \frac {\mu _0 \sigma \omega D}{4}$
Question 10
A charged particle moving with constant velocity passes through a space without any change in velocity
Electric Field in the region is given by the E=a1i + a2j + a3k
Magnetic Field in the region is given by the B=b1i + b2j + b3k
Where a1, a2, a3, b1, b2, b3 are constant
Which of the following can be true about the region?
(a) a1=a2=a3=b1=b2=b3 =0
(b) a1=a2=a3 =0, b1≠0 b2≠0 b3≠0
(c) a1≠0 a2≠0 a3≠0, b1=b2=b3 =0
(d) a1≠0 a2≠0 a3≠0 b1≠0 b2≠0 b3≠0 Solution
The force on the moving charge is given by F =q( E+ vXB)
Now for the velocity to be constant F=0
From the above relation, F=0 will true when
a) E=0, B=0
b) E=0, vXB =0 when velocity is parallel with magnetic field so B≠ 0
c) Electric force qE is equal and opposite to magnetic force q(vXB) so that mean B≠ 0 and E≠0
So, a, b, d is true
Question 11
A charged particle of mass m and charge q moves with a constant velocity along the positive x direction v=a i.
it enters a region of Magnetic field which is directed towards positive z direction from x=a which is given by B = b k
Find the initial acceleration of the particle
(a) a=(-qab/m) j
(b) a=(-qab/m) j
(c) a=(-qa/mb) j
(d) none of these Solution
Force on the charged particle is given by F=q(v X B)
So, force F=q[(ai)X(bk)]
=-qab j a=F/m a=(-qab/m) j
Question 12
Find the radius of the circular path which the particle moves
(a) mb/qa
(b) ma/q
(c) mab/q
(d) None of these Solution
We know that
$\frac {mv^2}{r}=qvB$
Or
r=mv/qB
=ma/qB
Question 13
Which of the following is true about the motion of the particle?
(a) Force will always be perpendicular to the velocity
(b) The magnitude of the velocity remains constant
(c) velocity vector and magnetic field vector remains perpendicular to each other during the motion
(d) KE of the particle remains constants Solution
All are true
Question 14
The frequency of the rotation
(a) depends on the value a
(b) depends on the value b
(c) depends on the value a and b both
(d) Does not depend on a and b both Solution
Angular velocity is given by
=v/r=qB/m
Frequency
=qB/2πm
=qb/2πm
So depends on b
Question 15
Four circular coil( A,B,C,D) having radius r,2r,3r,4r having numbers of turns 4n,3n,2n,n and carrying the same current are there. Match the Column Column X
(P) which coil will produce maximum Magnetic induction at the center
(Q) which coil will produce minimum Magnetic induction at the center
(R) Which coil has maximum magnetic moment per turn
(S) Which coil has minimum magnetic moment per turn Column Y
(E) A
(F) B
(G) C
(H) D Solution
Magnetic induction at the center of the coil is given by
$B=\frac {\mu _0 In}{2r}$
Here in this case $\frac {\mu _0I}{2}$ is a constant quantity
So, B =kn/r
So
B1 =k4n/r
B2 =k3n/2r
B3 =k2n/3r
B4 =kn/4r
So, coil A has maximum induction at the center
And coil D has minimum induction at the center
Now magnetic induction per unit coil
B=K/r
So Again, coil A has maximum induction at the center
And coil D has minimum induction at the center
Question 16
An infinite long thin cylinder shell of radius R is carrying current I along the axial direction.
Which of the following is true?
(a) B =0 for r < R
(b) B =k/r for r > R
(c) B=kr for r > R
(d) none of the above
Where $k= \frac {\mu _0 I}{2\pi }$ Solution
Let assume a circular path of radius r. By symmetry magnetic induction will be same at all the points and it will be acting tangential to the circular path
Applying ampere circuital law for outer loop r > R
$\oint B.dL=\mu _0I$
$B.2\pi r= \mu _0 I$
$B=\frac {\mu _0 I}{2\pi r}$
Applying ampere circuital law for inner loop r < R
$\oint B.dL=\mu _0I$
$B.2\pi r=0$
Paragraph Based Questions
(A) A particle of mass m and electric charge q is in a Magnetic field which varies as
$\mathbf{B}=\frac {\mu _0}{4 \pi}\frac {k}{r^3} \mathbf{r}$
Where k is constant and r is the radius vector from origin
Question 17
If the particle has velocity vector v, position vector r .Find the acceleration of the particle Solution
Question 18
which of the following is true
(a) $\mathbf{a}.\mathbf{v}=0$
(b) Speed v=|v| is constant in the motion
(c) $\mathbf{a}.\mathbf{r}=0$
(d) $m(\mathbf{r} \times \mathbf{v}) -q \mathbf{B} r^2$ Vector quantity is a constant of the motion Solution
From previous questions
$\mathbf{a} \perp \mathbf{v}$, So $\mathbf{a}.\mathbf{v}=0$
$\mathbf{a} \perp \mathbf{r}$, So $\mathbf{a}.\mathbf{r}=0$
Now
$\mathbf{a}.\mathbf{v}= \frac {1}{2} \frac{d}{dt} (\mathbf{v}.\mathbf{v}) = \frac {1}{2} \frac{d}{dt} (v^2) = v\frac {dv}{dt}$
Now $\mathbf{a}.\mathbf{v}=0$
So,
$v\frac {dv}{dt}=0$
$\frac {dv}{dt} =0$
or Speed v=|v| is constant in the motion
Now let's assume
$Q= m(\mathbf{r} \times \mathbf{v}) -q \mathbf{B} r^2$
Substituting the values of B
$Q= m(\mathbf{r} \times \mathbf{v}) -\frac {\mu _0}{4 \pi}\frac {k}{r} \mathbf{r}$
If this value, is constant of the motion ,then differentiation of it with respect to time should be zero
Hence all the options are correct
(B) A charge particle Q of mass m at rest is released from origin at t=0 in the xyz plane.
A constant magnetic field exists in the plane
$\mathbf{B}= B\mathbf{i}$
A constant electric field also exists in the plane
$\mathbf{E}= E\mathbf{k}$ Hint
It is given equation of following form gives following result
Question 19
The force acting on the particle at any point of time will be of the form
$\mathbf{F}= a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$
Which of the following is true about this motion?
(a) a is zero constant through out the motion
(b) b varies with the motion and it becomes zero at some points
(c) c varies with the motion and it never becomes zero
(d) b varies with the motion and it never become zero Question 20
Find velocity of the particle as a function of time t
Question 21
Find position vector of the particle as a function of time t Solution 19-21
Initially the particle is at rest, so magnetic force is zero and electric forces accelerates the charge in z-direction . As it pick up speed, a magnetic force develops which will pull the charge to the right.
$\mathbf{r}= x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$
Similarly, velocity vector at any time
$\mathbf{v}= v_x\mathbf{i} + v_y\mathbf{j} + v_z\mathbf{k}$
Now applying Newton law and lorentz law on the particle
$m\mathbf{a}= QE\mathbf{k} + Q(\mathbf{v} \times \mathbf{B})$
Substituting the value of B and v
$m\mathbf{a}= QE\mathbf{k} + QBv_z\mathbf{j} -QBv_y \mathbf{k}$ --(0)
or
$m(a_x\mathbf{i} + a_y\mathbf{j} + a_z\mathbf{k})=QE\mathbf{k} + QBv_z\mathbf{j} -QBv_y \mathbf{k}$
Therefore
$m \frac {d^2 x}{dt^2} =0$
which gives x=0 and $\frac {dx}{dt}=0$ always during the motion
$m \frac {d^2 y}{dt^2} =QB \frac {dz}{dt}$
$m \frac {d^2 z}{dt^2} =QE -QB \frac {dy}{dt}$
Let $ \omega = \frac {QB}{m}$
Then
$\frac {d^2 y}{dt^2} =\omega \frac {dz}{dt}$ --(1)
$ \frac {d^2 z}{dt^2} =\omega(\frac {E}{B} - \frac {dy}{dt})$ --(2)
Differentiating equation (1)
$ \frac {d}{dt} \frac {d^2 y}{dt^2} =\omega \frac {d^2z}{dt^2}$
substituting from (2)
$\frac {d}{dt} \frac {d^2 y}{dt^2}=\omega ^2 (\frac {E}{B} - \frac {dy}{dt})$
Integrating the above
$\frac {d^2 y}{dt^2} =\omega ^2 (\frac {E}{B}t - y)$
Therefore
$y= c_1 \cos \omega t + c_2 \sin \omega t + \frac {E}{B} t + c_3$ -(3)
Similarly Differentiating equation (2)
$ \frac {d}{dt} \frac {d^2 z}{dt^2} = -\omega \frac {d^2y}{dt^2}$
substituting from (1)
$ \frac {d}{dt} \frac {d^2 z}{dt^2} =-\omega ^2 \frac {dz}{dt}$
Integrating the above
$\frac {d^2 z}{dt^2} =-\omega ^2 z$
Therefore
$z= c_4 \cos \omega t + c_5 \sin \omega t + c_6$ -(4)
Now we know that
$y(0)=0$ and $(\frac {dy}{dt})_{t=0}=0$
$0=c_1 + c_3$
Differentiating equation (3) and at t=0
$0=\omega c_2 + \frac {E}{B}$
or
$c_2 = \frac {E}{B \omega}$
Differentiating the equation (3) again
$\frac {d^2 y}{dt^2}= -c_1 \omega ^2 \cos \omega t -c_2 \omega ^2 \sin \omega t$
Now at t=0 ,there is no acceleration along y direction
So $c_1$=0
And $c_3$=0
Hence
$y=\frac {E}{B \omega} ( \omega t - \sin \omega t)$ -(5)
Also
z(0)=0 and $(\frac {dz}{dt})_{t=0}=0$
From equation (4)
$0=c_4+c_6$
Differentiating equation (4) and at t=0
$0 =c_4 \omega$
or $c_5=0$
Differentiating the equation (4) again
$\frac {d^2 z}{dt^2}= -c_4 \omega ^2 \cos \omega t -c_5 \omega ^2 \sin \omega t$
Now $c_5=0$
And also t=0 , acceleration =QE/m
$\frac {QE}{m} = -c_4 \omega ^2$
or
$c_4 = - \frac {E}{\omega B}$
So
$c_6= \frac {E}{\omega B}$
Hence
$z= \frac {E}{\omega B}( 1- \cos \omega t)$ -(6)
From (5) and (6)
$\mathbf{r}=\left [ \frac {E}{B \omega} ( \omega t - \sin \omega t) \mathbf{j} + \frac {E}{\omega B}( 1- \cos \omega t) \mathbf{k }\right ]$
Differentiating
$\mathbf{v}=\left [ \frac {E}{B} ( 1 - \cos \omega t) \mathbf{j} + \frac {E}{B} \sin \omega t) \mathbf{k }\right ]$
Also from equation (0)
$m\mathbf{a}= QE\mathbf{k} + QBv_z\mathbf{j} -QBv_y \mathbf{k}$
So,
a is zero constant
b varies with the motion and it becomes zero at some points
c varies with the motion and it never becomes zero
Question 22
If the particle starts with the velocity at t=0
$\mathbf{v_0}=\frac {E}{B} \mathbf{j}$
Find out which of the following is true
(a) The particle moves in a straight line along y axis
(b) The velocity of the particle does not change during the motion
(c) The velocity vector will vary like
$\mathbf{v}=\frac {E}{B} \mathbf{j} + \frac {E}{B} \sin \omega t \mathbf{k}$ where $\omega =\frac {QB}{m}$
(d) The particle moves like a cycloid Solution
At t=0
Two forces are existing
(a) magnetic forces as velocity is present
(b) electric force due to charge particle
So, net force is zero and
particle will keep moving in j direction with constant velocity
Multiple Choice Questions
Question 23
A parabolic section of wire is located in the XY plane and caries current i. A uniform magnetic field exists throughout the plane and it is given by
$\mathbf{B}=B_0( \mathbf{i} + \mathbf{j})$
The equation of the parabola is
$y=4x^2$
Calculate the total force on the wire between origin and the point P where x=1
(a) $(4IB_0) \mathbf{k}$
(b) $(-4IB_0) \mathbf{k}$
(c) $(3IB_0) \mathbf{k}$
(d) $(-3IB_0) \mathbf{k}$ Solution
Question 24
A uniformly positively charged disk of mass $M_0$ whose total charge has magnitude Q and whose radius is R rotates with constant angular velocity of magnitude $\omega$
Which one of the following is true
(a) Angular momentum of the disk is $\frac {1}{2}M_0 R^2 \omega$
(b) The ratio of Magnetic moment and angular momentum is $\frac {Q}{2M_0}$
(c) The angular momentum and Magnetic moment direction are antiparallel
(d) None of these Solution
Question 25
The current density in the wire varies with r according to the formula
$J=r^2$ for r< a
$J=0$ for r > a
(a) The magnetic field is proportional to 1/r for r > a
(b) The magnetic field at r=a is given by
$B = \frac {\mu _0 a^3}{4}$
(c) The magnetic field is proportional to $r^4$ for r < a
(d) The magnetic field is proportional to $r^3$ for r < a Solution
Correct Answer are (a) (b) and (d)
Let us Choose a circular path centered on the conductor and apply ampere law to find the magnetic field
For circular path r > a
$B(2 \pi r) = \mu _0 I$
Now
$I=\int JdA = \int_{0}^{a} r^2(2 \pi r dr) =\frac {\pi a^4}{2}$
So
$B= \frac {\mu _ a^4}{4r}$
For circular path r < a
$B(2 \pi r) = \mu _0 I$
Now
$I=\int JdA = \int_{0}^{r} r^2(2 \pi r dr) =\frac {\pi r^4}{2}$
So
$B= \frac {\mu _ r^3}{4}$
At r=a
$B = \frac {\mu _0 a^3}{4}$
Question 26
A proton is fixed at the origin and electron revolves about it in a circular path of radius R.
$M_p$ -> Mass of proton
$M_e$ -> Mass of electron
q -> Charge on electron and proton
Find the magnetic field at the proton
(a) $B= \frac {\mu _0 q^2}{4 \pi R^2 \sqrt {4 \pi \epsilon _0 M_e R}}$
(b) $B= \frac {\mu _0 q^2}{2 \pi R^2 \sqrt {4 \pi \epsilon _0 M_e R}}$
(c) $B= \frac {\mu _0 q^2}{2 \pi R^2 \sqrt {4 \pi M_e R}}$
(d) $B= \frac {\mu _0 q^2}{8 \pi R^2 \sqrt {4 \pi \epsilon _0 M_e R}}$ Solution
The proton and electron are attracted by the Coulomb force
$F= \frac {q^2}{4 \pi \epsilon _0 R^2}$
This force will serve as the centripetal force for the circular motion, thus
$\frac {q^2}{4 \pi \epsilon _0 R^2}= M_e \omega ^2 R$
$\omega ^2= \frac {q^2}{4 \pi \epsilon _0 R^3 M_e}$
Now the electron is equivalent to a current loop
$I= \frac {\omega}{2 \pi} q$
The field for the circular loop at the center is given by
$B= \frac {\mu _0 I}{2 R} = \frac { \mu _0 \omega q}{4 \pi R}$
or
$B= \frac {\mu _0 q^2}{4 \pi R^2 \sqrt {4 \pi \epsilon _0 M_e R}}$
Paragraph Based Questions
(C) A positive charge particle of mass m and charge q is released from rest at x =-a ,y=0 and z=0 in the XYZ plane.
Electric and magnetic field exists in the XYZ plane as per below functions
$\mathbf{E}= E_0 \mathbf{i}$ for $x \leq 0$
$\mathbf{B}= -B_0 \mathbf{k}$ for $x \geq 0$
Question 27
Find the velocity of the particle when it reaches origin
(a) $v= \sqrt { \frac {2qE_0a}{m} }$
(b) $v= \sqrt { \frac {2qB_0a}{m} }$
(c) $v= \sqrt { \frac {2qE_0a}{B_0m} }$
(d) $v= \sqrt { \frac {2qB_0a}{E_0m} }$ Solution
$\mathbf{E}= E_0 \mathbf{i}$
So, charge particle will accelerate in the electric field and work done by the electric field will be converted into kinetic energy
So, when the particle reaches at origin
$\frac {1}{2} mv^2 = qE_0 a$
Or
$v= \sqrt { \frac {2qE_0a}{m} }$
Question 28
Find the coordinates of particle when it re-enters the electric field region
(a) $x = \sqrt { \frac {8mE_0 a}{qB_0^2}}$ ,$y=0$ ,$z=0$
(b) $x = 0$ ,$y=0$,$z=\sqrt { \frac {8mE_0 a}{qB_0^2}}$
(c) $x = \sqrt { \frac {2mE_0 a}{qB_0^2}}$ ,$y=\sqrt { \frac {8mE_0 a}{qB_0^2}}$ ,$z=0$
(d) $x = 0$ ,$y=\sqrt { \frac {8mE_0 a}{qB_0^2}}$ ,$z=0$ Solution
When the charge enters in Magnetic field, it will be acted by the magnetic force given by
$\mathbf{F}= q (\mathbf{v} \times \mathbf{B})=q[(v\mathbf{i}) \times (-B_0 \mathbf{k}) ]=qvB_0 \mathbf{j}$
The magnetic force acts perpendicular to the velocity. So it won't do any work and speed remains constant and particle moves in a circle of Radius R given by
$qvB_0^2 = \frac {mv^2}{R}$
or
$R= \frac {mv}{qB_0}=\sqrt { \frac {2mE_0 a}{qB_0^2}}$
The particle will move in semi circle path and reenter the electric field region at distance 2R from entry point
So, coordinates are
$x = 0$ ,$y=\sqrt { \frac {8mE_0 a}{qB_0^2}}$ ,$z=0$
Match the Column
Question 29
M is defined as
$M= \frac {B}{\mu _0}$ Column A
(P) M at the center of square loop which carries a steady current I and let a be the distance from the center to the side
(Q) M at the center of regular 6 sided polygon which carries a steady current I and let a be the distance from the center to the side
(R) M at the center of n sided polygon polygon which carries a steady current I and let a be the distance from the center to the side
(S) M at the center of n sided polygon polygon in the limit n-> 8 which carries a steady current I and let a be the distance from the center to the side Column B
(A) $\frac {I}{2a}$
(B) $\frac {I \sqrt 2}{ \pi a}$
(C) $\frac {3I}{ 2\pi a}$
(D) $\frac {nI}{ 2\pi a} \sin \frac {\pi}{n}$
(E) $\frac {I}{ 2\pi a} \sin \frac {\pi}{n}$ Solution
Magnetic Field at a distance s from current carrying wire is given by
$B= \frac {\mu_0 I}{4 \pi s} ( \sin \theta _2 - \sin \theta _1)$ Square
For any side
$\theta _2 = - \theta _1 = 45^0$
So, Magnetic field at center due to one side
$B= \frac {\mu_0 I}{4 \pi a} \frac {2}{\sqrt 2}=\frac {\mu_0 I}{2 \pi a \sqrt 2}$
Total magnetic Field due to four sides
$B= 4 \times \frac {\mu_0 I}{2 \pi a \sqrt 2}= \frac {\mu_0 I \sqrt 2}{ \pi a }$
or
$\frac {B}{\mu _0} = \frac { I \sqrt 2}{ \pi a }$ 6-sided Polygon
For any side
$\theta _2 = - \theta _1 = 30^0$
So, Magnetic field at center due to one side
$B= \frac {\mu_0 I}{4 \pi a} \frac {2}{ 2}=\frac {\mu_0 I}{4 \pi a }$
Total magnetic Field due to six sides
$B= 6 \times \frac {\mu_0 I}{4 \pi a }= \frac {3\mu_0 I }{ 2 \pi a }$
or
$\frac {B}{\mu _0} = \frac {3 I }{ 2\pi a }$ n sided Polygon
For any side
$\theta _2 = - \theta _1 = \frac {pi}{n}$
So, Magnetic field at center due to one side
$B= \frac {\mu_0 I}{4 \pi a} \times 2 \times \sin \frac {\pi}{ n}=\frac {\mu_0 I}{2 \pi a } \sin \frac {\pi}{ n}$
Total magnetic Field due to n sides
$B= n \times \frac {\mu_0 I}{2 \pi a } \sin \frac {\pi}{ n}= \frac {n\mu_0 I}{2 \pi a } \sin \frac {\pi}{ n}$
or
$\frac {B}{\mu _0} = \frac {nI}{2 \pi a } \sin \frac {\pi}{ n}$
When $n \rightarrow \infty$
$\sin \frac {\pi}{n} = \frac {\pi}{n}$
$\frac {B}{\mu _0} = \frac {nI}{2 \pi a } \sin \frac {\pi}{ n}$=\frac {nI}{2 \pi a } \frac {\pi}{n}= \frac {I}{2a}$
Question 30
A long straight wire along the z axis carries a current I in negative x direction. The magnetic field vector B at a point having coordinates (x,y) in the z=0 plane is Solution
Question 31
A thick slab extending from z=-b to z=+b carries a uniform current density K and direction of current is positive x axis. Which of the following is true?
(a) Magnetic field points in negative y axis for z > 0
(b) Magnetic field points in negative y axis for z < 0
(c) Magnetic field points in positive y axis for z > 0
(d) Magnetic field points in positive y axis for z < 0 Solution
It is clear from Fleming right hand rule, that
Magnetic field points in negative y axis for z > 0 and
Magnetic field points in positive y axis for z < 0
Question 32
Two coaxial plane coil of radius r and separated by the distance r. The current I flows in same direction in both the coil
Which of the following statement is true?
(a) The magnetic Field on the axis at a point midway between the center is $\frac{8\mu_0I}{5r\sqrt5}$
(b) the magnetic field due to one coil on the axis decrease as the distance increases
(c) The magnetic field between the coil is relatively uniform between the coils.
(d) None of these Solution
Magnetic Field due to coil on the axis is given by
$B=\frac{\mu_0Ir^2}{2(r^2+x^2)^{3/2}} $ ---(1)
For x=r/2
$B=\frac{\mu_0I}{2r(5/4)^{3/2}}$
Magnetic due to both the coil
$B=\frac{\mu_0I}{r(5/4)^{3/2}}=\frac{8\mu_0I}{5r\sqrt5}$
From equation (1),it is clear option (b) is correct
Now For coil 1, magnetic field decrease as the distance is increase, Similar is the case of that. So their sum remains almost uniform
$B=\frac{\mu_0Ir^2}{2(r^2+x^2)^{3/2}}+\frac{\mu_0Ir^2}{2(r^2+(r-x)^2)^{3/2}} $
Now when x try decrease the first quantity, It increase the second quantity. So sum is relatively same between the coil
Question 33
Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper, as shown. The variation of the magnetic field B along the line XX' is given by Solution
If the current flows out of the paper, the magnetic field at points to the right of the wire will be upwards and to the left will be downward.
Now magnetic field at C, is zero.
The field in the region BX' will be upwards (+ve) because all points lying in this region are to the right of both the wires.
Similarly, magnetic field in the region AX will be downwards (- ve). The field in the region AC will be upwards (+ve) because points are closer to A compared to B. Similarly magnetic field in region BC will be downward (-ve)
Graph (b) satisfies all these condition
Question 34
A current carrying loop is placed in a uniform magnetic field in four different orientations I, II, III and IV as shown in figure. Arrange them in decreasing order of potential energy.
(a) I > III > II > IV
(b) I > IV > II > III
(c) I > II > III > IV
(d) III > IV > I > II Solution
As we know that, potential energy of a magnet in a magnetic field
$U=-m.B$
$=-mB cos \theta$
where, m= magnetic dipole moment of the magnet
B= magnetic field
Case I
$\theta=180^0$
$U_1=-mB \cos 180 =mB$
Case II
$\theta=90^0$
$U_2 =0
Case III
$\theta < 90$
Therefore
$U < 0$
Case IV
$\theta > 90$
$U > 0 and U < mb$
Therefore, decreasing order of PE is
I > IV > II > III
Question 35
In a cyclotron, a charged particle
(a) undergoes acceleration all the time.
(b) speeds up between the dees because of the magnetic field.
(c) speeds up in a dee.
(d) slows down within a dee and speeds up between dees. Solution
(a)
Question 36
An infinitely long hollow conducting cylinder with inner radius r/2 and outer radius R carries a uniform current density along its length . The magnitude of the magnetic field ,|B| as a function of the radial distance r from the axis is best represented by Solution
for x < R/2
B=0
for x > R/2 and x < R
$|B| = \frac {\mu _0 J}{2x} (x^2 - \frac {R^2}{4})$
For x > R
$|B|=\frac {3 \mu_0 JR^2}{8x}$
So (b) is the correct answer
