Question: Q. 13. A rectangular loop of wire of size $4 \mathrm{~cm} \times 10 \mathrm{~cm}$ carries a steady current of 2 A. A straight long wire carrying 5 A current is kept near the loop as shown. If the loop and the wire are coplanar, find :
(i) the torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current carrying wire.
A [Delhi I, II, III 2012]
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Solution:
Ans. (i) Torque on the loop
$$ \tau=M B \sin \theta $$
As $M$ and $B$ are parallel, $\theta=0$
$$ \text { Therefore, } \quad \tau=0 $$
(ii)Force acting on the loop
$$ \begin{align*} & \begin{align*} &|F|=\frac{\mu_{0} I_{1} I_{2}}{2 \pi} l\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right) \ &=2 \times 10^{-7} \times 2 \times 2 \times 10^{-1}\left(\frac{1}{10^{-2}}-\frac{1}{5 \times 10^{-2}}\right) \mathrm{N} \quad 1 / 2 \ &=\frac{8 \times 10^{-8}}{10^{-2}}\left(1-\frac{1}{5}\right) \mathrm{N} \ &=8 \times 10^{-6}\left(\frac{4}{5}\right) \mathrm{N} \ &=6.4 \times 10^{-6} \mathrm{~N} \end{align*} \ & \text { Direction : Towards conductor / Attractive. } 1 / 2 \ & 1 / 2 \end{align*} $$
[CBSE Marking Scheme 2012]
[A] Q. 14. Two long straight parallel conductors carrying steady currents $I_{1}$ and $I_{2}$ are separated by a distance ’ $d$ ‘. Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.
U [O.D. I, II, III 2012, 11]
Ans. Try yourself, Similar to Q. 12, Short Answer Type II
The nature of the force is repulsive for currents in opposite direction and attractive when currents flow in the same direction.
[CBSE Marking Scheme 2011-12]
