CBSE-XI-Physics
35: Magnetic Field due to a Current
- #11Two 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 µT. If the direction of i2 is reversed, the field becomes 30 µT. The ratio i1/i2 is
(a) 4
(b) 3
(c) 2
(d) 1digAnsr: cAns : (c) 2
The magnetic field due to the current-carrying long, straight wire at point a is given by
`` B=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}i}{2\,\mathrm{\,\pi \,}a}``
When both the wires carry currents i1 and i2 in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.
`` B\text{'}=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}}{2\,\mathrm{\,\pi \,}a}-\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{2}}{2\,\mathrm{\,\pi \,}a}=10\,\mathrm{\,\mu T\,}..\left(1\right)``
Here, a is the distance of the midpoint from both the wires.
When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.
`` B\text{'}\text{'}=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}}{2\,\mathrm{\,\pi \,}a}+\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{2}}{2\,\mathrm{\,\pi \,}a}=30\,\mathrm{\,\mu T\,}..\left(2\right)``
On solving eqs. (1) and (2), we get
`` {i}_{1}-{i}_{2}=10``
`` {i}_{1}+{i}_{2}=30``
`` \Rightarrow {i}_{1}=20,{i}_{2}=10``
`` \Rightarrow \frac{{i}_{1}}{{i}_{2}}=\frac{2}{1}=2``
Page No 249: