Magnetic Effects of Current

Q: Five long wires A, B, C,D, and E, each carrying current I are arranged to form edges of a pentagonal prism as shown in figure. each carries current out of the plane of paper.

(a) What will be magnetic induction at a point on the axis O , Axis is at a distance R from each wire ?

(b) What will be the field if current in one of the wires (say A) is switched off ?

(c) What if current in one of the wire (say A) is reversed ?

Solution :(a) Magnetic field due to one wire is $\displaystyle B = \frac{\mu_0}{4\pi}\frac{2 I}{R} $

Magnetic field at O due to all the five wires will be Zero .

(b) As total magnetic field is Zero . Hence magnetic field due to wire A = Magnetic field due to all four wires B , C , D & E

Hence Magnetic field due to all four wires will be $\displaystyle B = \frac{\mu_0 I}{2 \pi R} $  ; Towards Left as magnetic field due ti wire A is towards right .

(c) If current in wire A is reversed then net magnetic field at O will be

B = Magnetic field at O due to A + Magnetic field at O due to wire B , C , D & E

$\displaystyle B = \frac{\mu_0}{4\pi}\frac{2 I}{R} + \frac{\mu_0}{4\pi}\frac{2 I}{R} $ ; Towards Left .

$\displaystyle B = \frac{\mu_0 I}{\pi R} $ (Towards Left .)

Q: Two long wires carrying current I₁ and I₂ are arranged as shown in figure. the one carrying current I₁ is along is the x-axis. The other carrying current I₂ is along a line parallel to the y-axis given by x = 0 and z = d. Find the force exerted at O2 because of the wire along the x-axis.

Solution : Magnetic field at O2 due to wire carrying current I₁ is

$\displaystyle B = \frac{\mu_0}{4 \pi} frac{2 I_1}{d}$ , and direction is $\hat{-j}$

$\displaystyle B = \frac{\mu_0}{2 \pi} frac{I_1}{d}$

Force acting on wire carrying current I₂ will be

$\displaystyle \vec{F} = I_2 (\vec{l} \times \vec{B})$

The angle between $\vec{l}$ and $\vec{B}$ is 180° because it is anti parallel .

Hence force exerted at O₂ because of the wire along the x-axis is Zero .

Q: A long straight wire carrying current of 25 A rests on a table as shown in figure. another wire PQ of length 1m , mass 2.5 g carries the same current in opposite direction . The wire PQ free to move up & down. To what height will PQ rise ?

Solution : Let PQ will rise to height ‘ h ‘ then

Magnetic field due to ling wire carrying current I at a height h will be

$\displaystyle B = \frac{\mu_0}{4 \pi} \frac{2 I}{h}$

Force acting on PQ is $\displaystyle F = I l B$

$\displaystyle F = I l (\frac{\mu_0}{4 \pi} \frac{2 I}{h}) $

And F = m g

$\displaystyle m g = \frac{\mu_0}{2 \pi} \frac{I^2 l }{h} $

$\displaystyle h = \frac{\mu_0}{2 \pi} \frac{I^2 l }{m g} $

Q: Three semi infinite mutually perpendicular conductors are joined at the origin O as shown in figure. A current 2I enters through the conductor along Z-axis towards the origin O and leaves through the other two as shown in the figure. What is the magnetic field at a point P with position vector $\hat{i} + \hat{j} + \hat{k}$

Solution : Consider the current which is along y-axis .

Consider a current element at a distance y from the origin. The co-ordinates of this current element are (0,y,0). Hence position vector of point P with respect to current element is $\vec{R} = \vec{r} – y \vec{j} = \hat{i} + (1-y)\hat{j} + \hat{k}$

$\displaystyle \vec{dB} = \frac{\mu_0}{4 \pi} \frac{I dy\hat{j} \times \hat{i} + (1-y)\hat{j} + \hat{k}}{[2 + (1-y)^2]^{3/2}}$

$\displaystyle \vec{dB} = \frac{\mu_0 I}{4 \pi} \frac{dy (\hat{i} – \hat{k})}{[2 + (1-y)^2]^{3/2}}$

Let 1 – y = 2 tanθ

– dy = 2 sec²θ dθ

$\displaystyle \vec{dB} = – \frac{\mu_0 I}{4 \pi} \frac{2 sec^2 \theta d\theta}{2 sec^3 \theta}(\hat{i}-\hat{k}) $

$\displaystyle \vec{B_y} = – \frac{\mu_0 I}{4 \pi} (\hat{i}-\hat{k}) \int_{\theta = tan^{-1/2}}^{-\pi/2} cos\theta d\theta$

Similarly we can find negative field due to the conductor along X-axis

$\displaystyle \vec{B_x} = – \frac{\mu_0 I}{4 \pi} (\hat{k}-\hat{j}) \int_{\theta = tan^{-1/2}}^{-\pi/2} cos\theta d\theta$

and the magnetic field at P due to the conductor along z axis as,

$\displaystyle \vec{B_z} = – \frac{\mu_0 I}{4 \pi} (\hat{j}-\hat{i}) \int_{\theta = tan^{-1/2}}^{-\pi/2} cos\theta d\theta$

Thus the net magnetic field at P is

$\displaystyle \vec{B} = \vec{B_x} + \vec{B_y} + \vec{B_z} $

Q: A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic field E and B are presented. E and B are parallel to each other. At time t = 0 the velocity v₀ of the particle is perpendicular to E. (Assume that its is always << c, the speed of light in vacuum). Find the velocity v of the particle at time t. You must answer in terms of t, q, m the vector v₀. E and B, and their magnitude v₀ , E and B.

Solution : Let velocity of the particle is v0 along x axis.

$\displaystyle \vec{F} = [q \vec{E} + q (\vec{v} \times \vec{B}) ]$

$\displaystyle \vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k} $

$\displaystyle \vec{F} = m ( \frac{dv_x}{dt} \hat{i} + \frac{dv_y}{dt} \hat{j} + \frac{dv_z}{dt} \hat{k}) $

$\displaystyle = q [E\hat{j} + (v_x \hat{i} + v_y \hat{j} + v_z \hat{k}) \times B \hat{j}] $

$\displaystyle m \frac{dv_x}{dt} = – q v_z B $

$\displaystyle m \frac{dv_y}{dt} = q E $

$\displaystyle m \frac{dv_z}{dt} = – q v_x B $

$\displaystyle m \frac{d^2v_x}{dt^2} = – q B \frac{dv_z}{dt} $

$\displaystyle m \frac{d^2v_y}{dt^2} = – q B \frac{dv_x}{dt} $

v_z = 0 ; v_y = 0 ; v_x = v₀ at time t = 0

$\displaystyle v_x = v_0 cos(\frac{qB}{m}) t $

$\displaystyle v_y = (\frac{qE}{m}) t$

$\displaystyle v_z = v_0 sin(\frac{qB}{m}) t $

velocity of the particle

$\displaystyle \vec{v} = v_0 cos(\frac{qB}{m}) t \frac{\vec{v_0}}{v_0} + (\frac{qE}{m}) t \frac{\vec{E}}{E} + v_0 sin(\frac{qB}{m}) t \frac{\vec{E} \times \vec{B}}{|\vec{E} \times \vec{B}|}$

Q: A long straight conductor carrying current I₁ is placed in the plane of ribbon at a distance ‘ a ‘ from the near edge of a ribbon of width ‘ b ‘ which carry current I₂ parallel to the wire. Find the force of attraction between the two.

Solution : dF (force of attraction) due to thin strip of thickness dx at a distance x

$\displaystyle dF = \frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{x}(\frac{dx}{b}) $

$\displaystyle F = \frac{\mu_0}{2 \pi} \frac{ I_1 I_2}{b} \int_{a}^{a+b} \frac{dx}{x} $

$\displaystyle F = \frac{\mu_0 I_1 I_2}{2 \pi b} ln(\frac{a+b}{a}) $

Q: Find the current in the conductor PQ so that it remains in equilibrium under its own weight and the magnetic force. If direction of current is reversed then what would be the initial acceleration of the rod ? Mass of the rod is m. (Friction is assumed to be absent)

Solution : Magnetic force on the conductor is equal to Fm = B I l . This force is along horizontal. Direction of I should be from Q to P. In this case its component F_m cosθ will balance mg sinθ to keep it in equilibrium.
Since the rod can not move along perpendicular to the plane of the rails.

Hence for equilibrium F_m cosθ = m g sinθ

I l B = m g tanθ

$\displaystyle I = \frac{m g tan\theta}{ B l}$

On reversing the direction of current acceleration of the rod will be

$\displaystyle a = \frac{F_m cos\theta + m g sin\theta}{m}$

$\displaystyle a = \frac{I l B cos\theta + m g sin\theta}{m}$

a = 2 g sinθ

Q: A long thin walled hollow cylinder of radius r is carrying a current I. A very long current carrying straight conductor is passing through the axis of the hollow cylinder, carrying a current i₀. Find the tension per unit length developed in the hollow cylinder due to the interaction of the straight current carrying conductor.

Solution : Considering the magnetic forces between io and the differential current di we obtain

$\displaystyle dF = \frac{\mu_0 i_0}{2 \pi r }(di) $ $dl = \frac{i_0 d\theta}{2\pi}$

where dF is the force per unit length. Since the elementary portion is in equilibrium

2T sin (dθ/2) = dF (T is the tension per unit length)

⇒ Tdθ = dF [as dθ is very small, sin(dθ/2) ≅  dθ/2]

$\displaystyle T d\theta = \frac{\mu_0 i_0}{2 \pi r} (\frac{i d\theta}{2\pi}) $

$\displaystyle T = \frac{\mu_0 i i_0}{4 \pi^2 r} $

Q: Three very long straight current carrying conductors are placed parallel to each other as shown in the figure. The conductors 1 & 3 are fixed where as conductor 2 is free to move. If the conductor 2 is pulled toward right through a very small distance x, find the net force acting on it and the angular frequency of the resulting oscillation.

Solution : The net (restoring) force acting on the conductor after displacing it through a very small distance x is given as

F = | F₁ – F₂ |

$\displaystyle F = \frac{\mu_0 I i_0}{2\pi (d-x)} – \frac{\mu_0 I i_0}{2\pi (d+x)}$

$\displaystyle F = \frac{\mu_0 I i_0}{2\pi}[ \frac{(d+x)-(d-x)}{(d^2 – x^2)} ] $

For x ≤ d

$\displaystyle F = \frac{\mu_0 I i_0 x}{\pi d^2} $

$\displaystyle m \omega_{osc}^2 x = \frac{\mu_0 I i_0 x}{\pi d^2} $

$\displaystyle \omega_{osc} = \sqrt{\frac{\mu_0 I i_0}{\pi m d^2}} $

Q: Two concentric circular loops of radii R and r(R>>r) contain currents I and i respectively. Find the maximum torque and maximum angular acceleration of smaller loop, if it is free to rotate about y axis where as the bigger loop is fixed. (Neglect the inductance of the system). The mass of the smaller loop is m.

Solution : The magnetic induction due to the larger current loop at the center O is,

$\displaystyle B = \frac{\mu_0 I}{2 R}$

The magnetic moment of the smaller loop = μ = (π r²)i

The maximum torque experienced by the smaller loop

$\displaystyle \tau = \mu B = (\pi r^2 i ) (\frac{\mu_0 I}{2 R}) $

$\displaystyle \tau = \frac{\mu_0 I i r^2 \pi}{2 R} $

The maximum angular acceleration of the smaller loop

$\displaystyle \alpha = \frac{\tau}{I}$ ; Where I = Moment of Inertia

$\displaystyle \alpha = \frac{\frac{\mu_0 I i r^2 \pi}{2 R}}{m r^2}$

$\displaystyle \alpha = \frac{\mu_0 I i}{2 m R}$