Magnetics Ques 133
- Two long parallel wires carrying currents $2.5 A$ and $I$ (ampere) in the same direction (directed into the plane of the paper) are held at $P$ and $Q$ respectively such that they are perpendicular to the plane of paper. The points $P$ and $Q$ are located at a distance of $5 $ $m$ and $2 $ $m$ respectively from a collinear point $R$ (see figure).
(1990, 8M)
(a) An electron moving with a velocity of $4 \times 10^{5} m / s$ along the positive $x$-direction experiences a force of magnitude $3.2 \times 10^{-20} N$ at the point $R$. Find the value of $I$.
(b) Find all the positions at which a third long parallel wire carrying a current of magnitude $2.5$ A may be placed, so that the magnetic induction at $R$ is zero.
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Answer:
Correct Answer: 133.(a) $4A$ (b) At distance $1 m$ from $R$ to the left or right of it, current is outwards if placed to the left and inwards if placed to the right of $6 R$.
Solution:
Formula:
Magnetic Force Acting On A Current Carrying Wire:
- Magnetic field at $R$ due to both the wires $P$ and $Q$ will be downwards as shown in figure.
Therefore, net field at $R$ will be sum of these two
$ \begin{aligned} B & =B _P+B _Q=\frac{\mu _0}{2 \pi} \frac{I _P}{5}+\frac{\mu _0}{2 \pi} \frac{I _Q}{2} \\ & =\frac{\mu _0}{2 \pi} (\frac{2.5}{5}+\frac{I}{2})=\frac{\mu _0}{4 \pi}(I+1) \\ & =10^{-7}(I+1) \end{aligned} $
(a) Net force on the electron will be,
$ F _m=B q v \sin 90^{\circ} $
or $\left(3.2 \times 10^{-20}\right)=\left(10^{-7}\right)(I+1)\left(1.6 \times 10^{-19}\right)\left(4 \times 10^{5}\right)$
or $\quad I+1=5 \Rightarrow I=4 A$
(b) Net field at $R$ due to wires $P$ and $Q$ is
$ B=10^{-7}(I+1) T=5 \times 10^{-7} T $
Magnetic field due to third wire carrying a current of $2.5 A$ should be $5 \times 10^{-7} T$ in upward direction so, that net field at $R$ becomes zero. Let distance of this wire from $R$ be $r$. Then,
$ \begin{aligned} \frac{\mu _0}{2 \pi} \frac{2.5}{r} & =5 \times 10^{-7} \text { or } \frac{\left(2 \times 10^{-7}\right)(2.5)}{r}=5 \times 10^{-7} m \\ \text { or } \quad r & =1 m \end{aligned} $
So, the third wire can be put at $M$ or $N$ as shown in figure.
If it is placed at $M$, then current in it should be outwards and if placed at $N$, then current be inwards.
BETA
