Magnetics 6 Question 29
32. A circular loop of radius $R$ is bent along a diameter and given a shape as shown in figure. One of the semicircles $(K N M)$ lies in the $x-z$ plane and the other one $(K L M)$ in the $y-z$ plane with their centres at origin. Current $I$ is flowing through each of the semicircles as shown in figure.
(2000, 10M)
(a) A particle of charge $q$ is released at the origin with a velocity $\mathbf{v}=-v _0 \hat{\mathbf{i}}$. Find the instantaneous force $\mathbf{F}$ on the particle. Assume that space is gravity free. (b) If an external uniform magnetic field $B _0 \hat{\mathbf{j}}$ is applied determine the force $\mathbf{F} _1$ and $\mathbf{F} _2$ on the semicircles $K L M$ and $K N M$ due to the field and the net force $\mathbf{F}$ on the loop.
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Answer:
Correct Answer: 32. (a) $\mathbf{F}=-\frac{\mu _0 q v _0 I}{4 R} \hat{\mathbf{k}}$ (b) $\mathbf{F} _1=\mathbf{F} _2=2 B I R \hat{\mathbf{i}}, \mathbf{F}=4 B I R \hat{\mathbf{i}}$
Solution:
- Magnetic field $(\mathbf{B})$ at the origin = magnetic field due to semicircle KLM + Magnetic field due to other semicircle KNM
$$ \begin{aligned} \therefore \quad \mathbf{B} & =-\frac{\mu _0 I}{4 R}(\hat{\mathbf{i}})+\frac{\mu _0 I}{4 R}(\hat{\mathbf{j}}) \\ \mathbf{B} & =-\frac{\mu _0 I}{4 R} \hat{\mathbf{i}}+\frac{\mu _0 I}{4 R} \hat{\mathbf{j}}=\frac{\mu _0 I}{4 R}(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \end{aligned} $$
$\therefore$ Magnetic force acting on the particle
$$ \begin{aligned} & \mathbf{F}=q(\mathbf{v} \times \mathbf{B})=q{\left(-v _0 \hat{\mathbf{i}}\right) \times(\hat{\mathbf{i}}+\hat{\mathbf{j}}) } \frac{\mu _0 I}{4 R} \\ & \mathbf{F}=-\frac{\mu _0 q v _0 I}{4 R} \hat{\mathbf{k}} \end{aligned} $$
(b) $\mathbf{F} _{K L M}=\mathbf{F} _{K M N}=\mathbf{F} _{K M}$ and $\mathbf{F} _{K M}=B I(2 R) \hat{\mathbf{i}}=2 B I R \hat{\mathbf{i}}$
$$ \mathbf{F} _1=\mathbf{F} _2=2 B I R \hat{\mathbf{i}} $$
Total force on the loop, $\mathbf{F}=\mathbf{F} _1+\mathbf{F} _2$ or $\mathbf{F}=4 B I R \hat{\mathbf{i}}$ NOTE If a current carrying wire ADC (of any shape) is placed in a uniform magnetic field $\mathbf{B}$.
Then, $\quad F _{A D C}=F _{A C}$ or $\left|F _{A D C}\right|=\hat{i}(A C) B$
From this we can conclude that net force on a current carrying loop in uniform magnetic field is zero. In the question, segments $K L M$ and $K N M$ also form a loop and they are also placed in a uniform magnetic field but in this case net force on the loop will not be zero. It would had been zero if the current in any of the segments was in opposite direction.
