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  • The current loop  formed by two circular segments of radii  and <im

The current loop pqrst formed by two circular segments of radii r_1(6cm) and r_2(8cm) with common centre at O, Carries a current I(3A) as shown in the figure. What is the magnetic field at the common centre 'O' and what will be the value of magnetic field at centre when \Theta =180^0.

Option: 1

2.73 \times 10^{-5} \mathrm{~T}


Option: 2

2.75 \times 10^{-5} \mathrm{~T}


Option: 3

2.95 \times 10^{-4} \mathrm{~T}


Option: 4

3.25 \times 10^{-5} \mathrm{~T}


Answers (1)

best_answer

Magnetic field at the centre 'O' due to curret through the whole current loop is

B=B_p_q+B_q_r+B_r_s directed normally downwards

    \begin{aligned} & =0+\frac{\mu_0}{4 \pi} \frac{I}{r_2} \theta+0+\frac{\mu_0}{4 \pi} \frac{I}{r_2}(2 \pi \theta \\ & =\frac{\mu_0 I}{4 \pi}\left[\frac{\theta}{r_2}+\frac{2 \pi-\theta}{r_1}\right] \end{aligned}

If \theta=180^{\circ}=\pi \text { radian then }

\begin{aligned} B & =\frac{\mu_0 I}{4 \pi}\left[\frac{\pi}{r_2}+\frac{2 \pi-\pi}{r_1}\right] \\ & =\frac{\mu_0 I}{4}\left[\frac{1}{r_2}+\frac{1}{r_1}\right] \end{aligned}

     \begin{aligned} & =\frac{\left(4 \pi \times 10^{-7}\right)}{4} \times 3 \times\left[\frac{100}{8}+\frac{100}{6}\right] \\ & =\frac{4 \pi \times 10^{-7}}{4} \times 3\left[\frac{300+400}{24}\right] \end{aligned}

     \begin{aligned} & =\frac{4 \times \frac{22}{7} \times 10^{-7}}{4} \times 3 \times \frac{700}{24 } \\ & =\frac{22}{28} \times 10^{-7} \times \frac{700}{2} \\ & =\frac{7700}{28} \times 10^{-7}=\frac{77}{28} \times 10^{-5} \mathrm{~T} \\ & =2.75 \times 10^{-5} \mathrm{~T} . \end{aligned}

Posted by

Divya Prakash Singh

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