Question

Two parallel long wires carry currents i1 and i2 with i1>i2.when the currents are in the same direction the magnetic field at a point Midway between the wires is 10 micro Tesla if the direction of i2 is reversed that field becomes 30 micro Tesla the ratio of i1 is to i2 is

Answer 1

$\huge{\bf{Solution:-}}$

The magnetic field due to an infinity long wire at a distance d from it is given as:

$\sf{B = \frac{\mu_{o}i}{2\pi d}}$

Hence the magnetic field due to the two current carrying wires in the same direction is gives us:

$\sf{B_{net} = B_{1}-B_{2}}$

Where $\sf{B_{1} = \frac{\mu_{o}i_{1} }{2\pi d}}$ where d is distance of midway point from either wire

while $\sf{B_{2}=\frac{\mu_{o}i_{2}}{2\pi d}}$

So, $\sf{B_{net}=\frac{\mu_{o}i_{1}}{2\pi d}-\frac{\mu_{o}i_{2}}{2\pi d}}$

or, $\sf{10 = \frac{\mu_{o}}{2\pi d}\;\;\;\;\;\; (i_{1} -i_{2})\;\;\;\;\;\;........(1)}$

Similarly, the magnetic field due to the two current carrying wires in the opposite direction is,

$\sf{B_{net} = B_{1}+B_{2}}$

or,  $\sf{B_{net}=\frac{\mu_{o}i_{1}}{2\pi d}+\frac{\mu_{o}i_{2}}{2\pi d}}$

$\bf{30 = \frac{\mu_{o}}{2\pi d}\;\;\;\;\;\; (i_{1} -i_{2})\;\;\;\;\;\;........(2)}$

Diving (1) by (2) we have,

$\sf{1/3 = \frac{i_{1}-i_{2}}{i_{1}+i_{2}}}$

or, $\sf{(i_{1}+i_{2})=3( i_{1}+i_{2})}$

or, $\sf{2i_{1} =4i_{4}}$

or, $\sf{\frac{i_{1}}{i_{2}}}=2$

$\boxed{\sf{\frac{i_{1}}{i_{2}}}=2}$

Answer 2

Answer:

The ratio of the currents is 2:1.

Explanation:

Given that,

The magnetic field

When the currents in same direction

$B_{1}=10\times10^{-6}\ T$

When the currents in opposite direction

$B_{1}=30\times10^{-6}\ T$

Two parallel long wires carry currents i₁ and i₂.

The magnetic field due to infinite long wire at a distance d from it is given by,

$B = \dfrac{\mu_{0}i_{1}}{2\pi d}$......(I)

The magnetic field when the currents are in same direction,

$B = B_{1}-B_{2}$

$B=\dfrac{\mu_{0}i_{1}}{2\pi d}- \dfrac{\mu_{0}i_{2}}{2\pi d}$

$B=\dfrac{\mu_{0}}{2\pi d}(i_{1}-i_{2})$ .....(I)

The magnetic field when the currents are in opposite direction,

$B' = B_{1}+B_{2}$

$B=\dfrac{\mu_{0}i_{1}}{2\pi d}+ \dfrac{\mu_{0}i_{2}}{2\pi d}$

$B=\dfrac{\mu_{0}}{2\pi d}(i_{1}+i_{2})$ .....(II)

Now, Divided equation (I) by equation (II)

$\dfrac{B}{B'}=\dfrac{\dfrac{\mu_{0}}{2\pi d}(i_{1}-i_{2})}{\dfrac{\mu_{0}}{2\pi d}(i_{1}+i_{2})}$....(III)

Put the value in the equation (III)

$\dfrac{10\times10^{-6}}{30\times10^{-6}}=\dfrac{i_{1}-i_{2}}{i_{1}+i_{2}}$

$\dfrac{1}{3}=\dfrac{i_{1}-i_{2}}{i_{1}+i_{2}}$

$(i_{1}+i_{2})=3(i_{1}-i_{2})$

$\dfrac{i_{1}}{i_{2}}=\dfrac{4}{2}$

$\dfrac{i_{1}}{i_{2}}=\dfrac{2}{1}$

Hence, The ratio of the currents is 2:1.