Magnetic field and magnetic effects of current Problems
Subjective Questions
Question 1 A current of 10A is flowing from east to west in a long wire kept in east-west direction. Find the magnetic field in a horizontal plane at a distance of (1) 10cm north (2) 20cm south from the wire; and in a vertical plane at a distance of (3) 30cm downward (4) 50cm upward. Solution
The magnetic field at a distance of R meter from along wire is
\(B=\frac{\mu _0}{2\pi}\frac{i}{R}=\left( 2\times 10^{-7} \right) \frac{i}{R}\) \(NA^{-1}m^{-1}\)
(1) The magnetic field in a horizontal plane at a distance of 10cm (0.10m) north from the wire
\(B_N=\frac{2\times 10^{-7}\times 10}{0.10}=2\times 10^{-5}\) \(NA^{-1}m^{-1}\)
The current in the wire is from east to west. So, according to the right hand palm rule no.1, the direction of the field at the point towards north will be downwards in a vertical plane.
(2) The magnetic field at a distance of 20cm (0.20m) south of the wire
\(B_S=\frac{2\times 10^{-7}\times 10}{0.20}=1\times 10^{-5}\) \(NA^{-1}m^{-1}\)
The direction of the field will be upward in the vertical plane.
(3) The magnetic field at a distance of 40cm (0.40m) from the wire downward in vertical plane is
\(B_L=\frac{2\times 10^{-7}\times 10}{0.40}=5\times 10^{-6}\) \(NA^{-1}m^{-1}\)
The field will be in horizontal plane pointing south.
(4) The magnetic field at a distance of 50cm (0.50m) above the wire in vertical plane is
\(B_U=\frac{2\times 10^{-7}\times 10}{0.50}=4\times 10^{-6}\)\(NA^{-1}m^{-1}\)
The field will be in horizontal plane pointing north.
Question 2 A short conductor of length 4cm is placed parallel to the long conductor of length 2m near to its centre at a distance of 2cm. The conductor carry currents of 2A and 5A respectively in opposite directions. Find the total force exert on the long conductor. Solution
Magnetic field at the location of the short conductor due to current in a long conductor is given by
\(B=\frac{\mu _0}{2\pi}\frac{i_1}{R}\)
Where R is the distance of the short conductor from the long conductor. The field is almost uniform over the whole length l (say) of the 'short' conductor. Therefore, the force experienced by the short conductor carrying current (say) is given by
\(F=i_2BL=\frac{\mu _0}{2\pi}\frac{i_1i_2l}{R}\)
Substituting the values:
\(F=\left( 2\times 10^{-7}NA^{-2} \right) \times \frac{\left( 5.0A \right) \left( 2.0 \right) \left( 4.0\times 10^{-2}m \right)}{\left( 2.0\times 10^{-2}m \right)}=4\times 10^{-6}N\)
By Newton's third law, the long conductor will also experience an equal repulsive force, \(4\times 10^{-6}N\), due to short conductor
Question 3 An electron-gun G emit electron of energy 2KeV travelling in the positive X-direction. The electron are required to hit the spot S where GS= 0.1m, and the line GS makes an angle of 60^{0} with X-axis, as shown in the figure. A uniform magnetic field B parallel to GS exist in the region outside the electronic-gun. Find the minimum value of B needed to make the electron hit S.
Given: m_{e} =9.1 x 10^{-31} kg , e=1.6 x 10^{-19 C}
Resolve velocity v of electron parallel and perpendicular to the magnetic field B, we have
\(v_{||}=v\cos 60^0\) And \(v_{\bot}=v\sin 60^0\)
Due to \(v_{\bot}\) the electron move along a circle perpendicular to B with uniform speed , where as due to \(v_{||}\) they move along B with constant speed\(v_{||}\). Hence the path of electron is a helix whose axis is GS. The radius of helix is
\(r=\frac{mv_{\bot}}{eB},\) \(\left[ \frac{mv_{\bot}^2}{r}=ev_{\bot}B \right] \)
Where m and e are mass and charges of electron.
The time-period of circular motion is \(T=\frac{2\pi r}{v_{\bot}}=\frac{2\pi r}{eB}\)
The time taken by the electron to cover the distance GS along B is
\(t=\frac{GS}{v_{||}}=\frac{0.1m}{v\cos 60^0\left( m/s \right)}=\frac{0.2}{v}\sec \)
In order that the electron hit the spot S, the time t should be an integral multiple of T, that is t=n(T), where n is an integral
Or
\[\frac{0.2}{v}=n\frac{2\pi r}{eB}
\\
\frac{B}{n}=\frac{2\pi r}{e}\frac{v}{0.2}=\frac{10\pi mv}{e}\]
For B to be minimum, n=1(minimum value of n=1)
\(\therefore B_{\min}=\frac{10\pi mv}{e}\) --(1)
The velocity v of the electron can be computed from their kinetic energy K (= .\(\left( = 2\times 10^{-3}\times 1.6\times 10^{-19}J \right) \)
\[\frac{1}{2}mv^2=K
\\
v=\sqrt{\frac{2K}{m}}\]
Substituting this of v in eq. (i), we get
\(B_{\min}=\frac{10\pi m}{e}\sqrt{\frac{2Km}{m}}=\frac{10\pi}{e}\sqrt{2Km}\)
\begin{aligned}
&=\frac{10\times 3.14\times \left[ 2\times \left( 2\times 10^3\times 1.6\times 10^{-19}J \right) \times \left( 9.1\times 10^{-31}kg \right) \right] ^{1/2}}{1.6\times 10^{-19}C}\\
&=4.74\times 10^{-3}T\\
\end{aligned}
Question 4 A particle of mass 1x 10^{-26 } kg and charge 1.6 x 10^{-19 } C travelling with velocity of 1.28 x 10^{6} m/s in +x direction enters a region having electric field E and a uniform magnetic field B such as E_{x} =E_{y} =0, E_{z} = -102.4 V/m and B_{x} =B_{z} =0, B_{y} =8x10^{-2} Wb/m^{2 } . The particle enters this region at a long time t=0. Determine the location (x, y and z co-ordinates) of the particle at t=5x10^{-6} s . If the electric field switched off at this instant (with the magnetic field still present), what will be the position of the particle at t=7.46 x10^{-6} s. Solution
The force on the particle due to the electric field is
\begin{aligned}
qE_z&=\left( 1.6\times 10^{-19}C \right) \times \left( -102.4\times 103 \right) V/m\\
qE_z&=-1.6384\times 10^{-14}N\\
\end{aligned}
in the �z direction.
The force on the particle due to the magnetic field is
\begin{aligned}
qvB_y&=\left( 1.6\times 10^{-19}C \right) \times \left( 1.28\times 10^6m/s \right) \times \left( 8\times 10^{-2}Wb/m^2 \right)\\
qvB_y&=1.6384\times 10^{-14}N\\
\end{aligned}
in the +z direction.
The two forces are at equal and opposite. So, the net force, and hence the net acceleration of the particle, is zero. The distance travelled in +x direction is
\begin{aligned}
x&=v\times \Delta t\\
x&=\left( 1.28\times 10^6m/s \right) \times \left( 5\times 10^{-6}s \right)\\
x&=6.4m\\
\end{aligned}
So, at time \(t=5\times 10^{-6}\sec \), the co-ordinate of the particle will be (6.4, 0, 0).
In the magnetic field \(B_y\) alone, the particle will move along the circumference of a vertical circle. If the radius of the circle be r, then
The distance travelled by the particle along the circumference of the circle from time \(t=5\times 10^{-6}sec \) to \(t=7.45\times 10^{-6} sec \) is
\(v\times \Delta t=\left( 1.28\times 10^6 \right) \times \left( 2.45\times 10^{-6} \right) =3.136m=\pi \,\,meter\)
Since the radius of the circle is 1m, this distance is half of the circumference, i.e. the particle has moved a distance of 2r=2meter along the +z direction. Hence the co-ordinates of the particle at time \(t=7.46\times 10^{-6}\sec \) are (6.4, 0, 2).
Question 5 In a certain region surrounding the origin of the coordinates, B=5 x10^{-4} a_{z} T and E=5a_{z} V/m. A proton ( q_{p} =1.602 x 10^{-19} C,m_{p} =1.6 x10^{-27} kg) enters in the field at the origin with a velocity u_{0} =2.5 x10^{5} a_{x} m/s. Describe the proton’s motion and give its position after three complete revolutions Solution
The initial force of the particle is
\(F_0=q\left( E+u_0\times B \right) =q_p\left( Ea_z-u_0Ba_y \right) \)
The z component (electric component) of the force is constant, and produces a constant acceleration in the z direction. Thus the equation of the motion in z direction is
\(z=\frac{1}{2}at^2=\frac{1}{2}\left( \frac{q_pE}{m_p} \right) t^2\)
The other (magnetic) component, which changes into \(-q_puBa\), produces circular motion perpendicular to the z-axis, with period
\(T=\frac{2\pi r}{u}=\frac{2\pi m_p}{q_pB}\)
The resultant motion is helical.
After three revolution, x=y=0 and
Question 6 Find an expression for magnetic field at the centre of a circular current carrying loop. Solution
Magnetic at the centre of the current carrying loop-
Consider a circular loop of radius R carrying current i. Let us first calculate the magnetic field B due to small element dl from the shown in fig.
\(\mathbf{d}l=Rd\theta \hat{\theta}_0\)
where \(\hat{\theta}_0\) is a unit vector in the \(\theta \) direction. Note that since \(dl\) is small means \(d\theta \) is small. The unit vector \(\hat{r}\) and \(\hat{\theta}_0\)are perpendicular to each other at all points along the path. Then, from equation, the field due to this element at the centre
\begin{aligned}
&=\frac{\mu _0i}{4\pi}\left( \frac{Rd\theta \hat{\theta}\times \hat{r}}{R^2} \right)\\
&=\frac{\mu _0}{4\pi}i\frac{Rd\theta}{R^2}\left( \hat{\theta}\times \hat{r} \right)\\
\end{aligned}
Hence, total B due to coil at centre O,
\(B=\int\limits_0^{2\pi}{\mathbf{d}B}=\left( \hat{\theta}\times \hat{r} \right) \frac{\mu _0}{4\pi}\int\limits_0^{2\pi}{\frac{id\theta}{R}}\)
\begin{aligned}
&=\left( \hat{\theta}\times \hat{r} \right) \frac{\mu _0}{4\pi}\int\limits_0^{2\pi}{\frac{id\theta}{R}}\\
&=\frac{\mu _0i}{2R}\hat{k}\\
\end{aligned}
\(\hat{\theta}\times \hat{r}=\hat{k}\) is a unit vector along z-axis because B is directed along the direction perpendicular to the containing \(\hat{\theta}_0\) and \(\hat{r}\) i.e. xy plane here. Hence, B is directed out of the plane of the paper, i.e. z-axis.
Question 7 A current of 1.0 A is flowing in the sides of an equilateral triangle of sides 4.5 x 10^{-2}m. Find the magnetic field at the centroid of the triangle. Solution
The magnitude of the magnetic field at the centroid O of the triangle due to side PQ is
\(\frac{\mu _0}{4\pi}\frac{i}{R}\left( \sin \varphi _1+\sin \varphi _2 \right) \)
where R is the perpendicular distance of PQ from O , and \(\varphi _1\) , \(\varphi _2\) are angles as shown. The field is perpendicular to the plane of the paper , directed downwards.Since the magnetic field due to each of the three sides is the same magnitude and direction , the magnitude of the resultant field at O is
\(B=3\frac{\mu _0}{4\pi}\frac{i}{R}\left( \sin \varphi _1+\sin \varphi _2 \right) \)
Here i=1.0A , \(\varphi _1 = \varphi _2\)=60° and R=(l/2)cot60 where l is the side of the triangle. Putting these values in above equation we find,
$B=4.0 \times 10^{-5}$T
Question 8 A charge q=40 µC moves with instantaneous velocity u=(5x10^{4}) j m/s through the uniform fields E=(6x10^{4}) (.52 i +.56j +.645 k) V/m , B=(1.7)(.693 i + .6 j + .4k) T find the magnitude and direction of the instantaneous force on q. Solution
From the Lorentz equation
$\mathbf{F}= q\mathbf{E} + q (\mathbf{v} \times \mathbf{B})$
From this,
\(F=\left[ \left( 2.61 \right) ^2+\left( 1.34 \right) ^2+\left( 0.81 \right) ^2 \right] ^{\frac{1}{2}}=3.04N\)
The direction cosines of F are
\(l=\frac{2.61}{3.04}=0.86\)
\(l=\frac{1.34}{3.04}=0.44\) \(l=\frac{-0.81}{3.04}=-0.27\)
Question 9
Find the force on each segment of the wire as shown below in the figure
if B=0.15T. Assume that the current in the wire is 5A. Solution
For each straight segment \(F=iL\times B\), where L is the direct line segment.
In section AB and DE, L and B are parallel, \(\sin \theta =0,\)and F=0.
In section BC, F=iBL= (5A)(0.16m)(0.15T)=0.12N, inward
and in section CD, \(F=(5A)(0.2m)(0.15T)\sin 65^0 =0.136N \), outward.
Question 10
A planar coil of 12 turns carries 15A. The coil is oriented with respect to the uniform magnetic field B=0.2i+0.3j-0.4k T such that its directed area is A=0.04i-0.05j+0.07k m^{2}. Find (a) the dipole moment of the coil (b)the potential energy of the in the given orientation, and (c)the angle between the positive normal to the coil and the field. Solution
(a) dipole moment m = niA=(15)(12)(0.04i-0.05j+0.07k)=7.2i-9j+12.6k A\(m^2\)
(b) Potential energy U= - m.B = -[(7.2)(.2) +(-9.0)(0.3) +(1.26)(-0.4)]=6.3J
(c) we know that U= - m.B = -mBcos? where ? is the desired angle between m and B.
\(\theta =\cos ^{-1}\left( \frac{U}{mB} \right) =133.24^{\circ}\)
Objective Type questions
Question 11
A wire in the form of semicircle of radius r lies on the top of a smooth table . A downward directed uniform magnetic field of magnitude B is confined to region above the dashed line in the below figure.
The ends of the semicircle are attached to springs C and D whose other ends are fixed. The current i is introduced by attaching a battery to the ends of the springs as shown above.
Let $T_1$ be the tension in the Spring C and $T_2$ be the tension in the spring D
Which of the one following is true
(a) $T_1 + T_2 = 2iBr$
(b) $T_1 - T_2 = 2iBr$
(c) $T_1 + T_2 = iBr$
(d) None of these Solution
There is a radially outward magnetic force dF on the element ds of the semicircle. The force has magnitude
$dF=iBds=iBrd \theta$
When these forces are summed over the entire semicircle,the components parallel to the diameter of the semicircle cancel. The net magnetic force is along OX of magnitude
$F_m=\int dF cos \theta =2iBr\int_{0}^{\pi/2} cos \theta d\theta=2iBr$
In order for the loop to be in equilibrium ,the springs C and D must exert a force equal and opposite to $F_m$. So
$T_1 + T_2 = 2iBr$
So (a) is the correct answer
Question 12
Find the torque which acts on the rectangular current loop as shown in below figure
The magnetic field is given by
$\mathbf{B}=B_0 \mathbf{i}$
(a) $iAB \cos \theta \mathbf{j}$
(b) $iAB \sin \theta \mathbf{j}$
(c) $ -iAB \cos \theta \mathbf{j}$
(d) $-iAB \sin \theta \mathbf{j}$ Solution
The area vector is perpendicular to Loop and it is making an angle $\theta$ with the x axis in the x-z plane,So
$\mathbf{A}= (A \cos \theta) \mathbf{i} - (A \sin \theta)\mathbf{k}$
Now Magnetic moment
$\mathbf{\tau} = i \mathbf{A} \times \mathbf{B}$
or
$ \mathbf{\tau}= i [(A \cos \theta) \mathbf{i} - (A \sin \theta)\mathbf{k}] \times [B_0 \mathbf{i}]=-iAB \sin \theta \mathbf{j}$
Question 13
Find the force on the triangular loop ABC as shown in figure near an infinite straight wire. Both the loop and the wire carry a steady current I.
Let us assume the co-ordinates system as shown in below fig
The force on the portion BC
$F_{BC}=\frac{\mu_0I^2a}{2\pi s}$ (upward)
Now magnetic Field vary at all the points on the portion AB and AC
Now Magnetic field at any point due to wire is given by
$B=\frac{\mu_0I}{2\pi s}\mathbf{k}$
Now
$d\mathbf{F}=I(d\mathbf{L} \times \mathbf{B})=I[(\mathbf{i} dx+\mathbf{j}dy+\mathbf{k}dz) \times \frac{\mu_0I}{2\pi y}\mathbf{k}]$
$=\frac{\mu_0I^2}{2\pi y}(-\mathbf{j}dx+\mathbf{i}dy)$
Now the x-component cancels the corresponding components from right hand side
So net force on the two side
$F=2\int{\frac{\mu_0I^2}{2\pi y}dx}$ (downward as negative sign)
Now to get the force we need to integrate from B to A
From the figure
$y=x\sqrt3$
So
$F=\int{\frac{\mu_0I^2}{\pi x\sqrt3}dx}$
$F=\frac{\mu_0I^2}{\pi\sqrt3}\int_{s/\sqrt3}^{s/\sqrt3+a/2}{\frac{1}{x}dx}$
Or
$F=\frac{\mu_0I^2}{\pi\sqrt3}ln{\frac{s/\sqrt3+a/2}{s/\sqrt3}}=\frac{\mu_0I^2}{\pi\sqrt3}ln{(}1+\frac{a\sqrt3}{2s})$
So Net force the triangular loop
$F=\frac{\mu_0I^2a}{2\pi s}-\frac{\mu_0I^2}{\pi\sqrt3}ln{(}1+\frac{a\sqrt3}{2s})$
Paragraph Based Questions
Find the magnetic dipole moment of the bookend shaped loop as shown in below figure. All the sides has length b and it carries a current I
Question 14
Find the magnetic dipole moment of the loop
(a) $Ib^2(\mathbf{j}+\mathbf{k})$
(b) $Ib^2(\mathbf{j}-\mathbf{k})$
(c) $Ib^2(\mathbf{i}+\mathbf{k})$
(d) $Ib^2(\mathbf{i}+\mathbf{j})$ Question 15
If a uniform magnetic field exists in the region
$B=B_0(\mathbf{i}+\mathbf{j})$
Find the torque acting on the loop
(a) $ib^2B_0(\mathbf{j}+\mathbf{k}-\mathbf{i})$
(b) $ib^2B_0(\mathbf{j}-\mathbf{k}+\mathbf{i})$
(c) $ib^2B_0(\mathbf{k}-\mathbf{i})$
(d) $ib^2B_0(\mathbf{j}-\mathbf{k}-\mathbf{i})$ Solution 14-15
This wire loop can be considered the superposition of the two plane square loops.The extra sides cancel each other when two are put together since the current flow in opposite direction
So magnetic dipole moment
$\mathbf{M}=I\mathbf{A}$
$\mathbf{M}=Ib^2\mathbf{j}+Ib^2\mathbf{k}$
Now torque is given by
$\tau=\mathbf{M} \times \mathbf{B}$
$=ib^2(\mathbf{j}+\mathbf{k}) \times B_0(\mathbf{i}+\mathbf{j})=ib^2B_0(\mathbf{j}-\mathbf{k}-\mathbf{i})$
Match the column
Question 16
A long straight metal rod of Radius $R_1$ has a very large hole of radius $R_2$ drilled parallel to the rod axis. The rod is carrying current I .The cross-section of the rod from the top is shown below
The distance between C and A is 3d Column A
(P) Magnetic field at Point A
(Q) Magnetic field at Point B
(R) Magnetic field at point C
Let's first calculate the current density in the rod
$J= \frac {I}{\pi (R_1^2 - R_2^2)}$
Now this system can be assumed as solid rod of $R_2$ having current $J \pi R_2^2$ one way and a second rod having current $J \pi R_1^2$ moving the other way.
Now we also know from Ampere law that
Magnetic field at the distance r from axis of the rod is given by
$B= \frac {\mu _0 JA}{2 \pi r}$
Where J is the current density in the rod
$B= \frac {\mu _0 J A}{2 \pi r}$
Now for point A( Point on the axis of bigger rod)
Magnetic field due to larger rod is zero and there exists a magnetic field due to smaller rod
So,
$B= \frac {\mu _0 J \pi R_2^2}{2 \pi d}=\frac {\mu _0 I R_2^2}{2 \pi d (R_1^2 - R_2^2)}$
Now for point B( Point on the axis of smaller rod)
Magnetic field due to smaller rod is zero and there exists a magnetic field due to bigger rod
$B= \frac {\mu _0 J \pi d^2}{2 \pi d}=\frac {\mu _0 I d}{2\pi (R_1^2 - R_2^2)}$
Now for point C( Point outside the bigger rod on the line joining the center of the rods)
Magnetic field due to smaller rod will be present and also magnetic field due to bigger rod will be present. And they will both oppose each other as direction of current is opposite
Magnetic field due to bigger rod
$B_1 = \frac {\mu _0 J \pi R_1^2}{2 \pi R}$
Magnetic field due to smaller rod
$B_2 = \frac {\mu _0 J \pi R_2^2}{2 \pi (R -d)}$
Net magnetic field
$B= \frac {\mu _0 J \pi R_1^2}{2 \pi R} - \frac {\mu _0 J \pi R_2^2}{2 \pi (R -d)}=\frac {\mu_0 I}{\pi ((R_1^2 - R_2^2)} [ \frac {R_1^2}{R} - \frac {R_2^2}{R-d}]$
Since R=3d
$B= \frac {\mu_0 I}{\pi ((R_1^2 - R_2^2)} [ \frac {R_1^2}{3d} - \frac {R_2^2}{3d-d}] = \frac {\mu _0 I(2R_1^2 - 3R_2^2)}{6 \pi d(R_1^2 - R_2^2)}$
Question 17
A charged particle would continue to move with a constant velocity in a region wherein,
(a) $\mathbf{E} = 0$, $\mathbf{B} \ne 0$
(b) $\mathbf{E} \ne 0$, $\mathbf{B} \ne 0$
(c) $\mathbf{E} \ne 0$, $\mathbf{B} = 0$
(d) $\mathbf{E} = 0$, $\mathbf{B} \ne 0$ Solution