JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 5)
Match List I with List II
| List I (Current configuration) |
List II (Magnitude of Magnetic Field at point O) |
||
|---|---|---|---|
| A. | _25th_January_Morning_Shift_en_5_1.png) |
I. | $${B_0} = {{{\mu _0}I} \over {4\pi r}}[\pi + 2]$$ |
| B. | _25th_January_Morning_Shift_en_5_2.png) |
II. | $${B_0} = {{{\mu _0}} \over {4 }}{I \over r}$$ |
| C. | _25th_January_Morning_Shift_en_5_3.png) |
III. | $${B_0} = {{{\mu _0}I} \over {2\pi r}}[\pi - 1]$$ |
| D. | _25th_January_Morning_Shift_en_5_4.png) |
IV. | $${B_0} = {{{\mu _0}I} \over {4\pi r}}[\pi + 1]$$ |
Choose the correct answer from the options given below :
A-III, B-IV, C-I, D-II
A-II, B-I, C-IV, D-III
A-III, B-I, C-IV, D-II
A-I, B-III, C-IV, D-II
Explanation
$A \rightarrow B_{0}=\frac{-\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{2 r}-\frac{\mu_{0} I}{4 \pi r}$
$$ B_{0}=\frac{\mu_{0} I}{2 \pi r}(\pi-1) \quad A \rightarrow \text { III } $$
$B \rightarrow B_{0}=\frac{\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{4 r}+\frac{\mu_{0} I}{4 \pi r}$
$$ B_{0}=\frac{\mu_{0} I}{4 \pi r}(\pi+2) \quad B \rightarrow I $$
$C \rightarrow B_{0}=\frac{\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{4 r}+0$
$B_{0}=\frac{\mu_{0} I}{4 \pi r}(\pi+1) \quad \quad \mathrm{C} \rightarrow \mathrm{IV}$
D. $\rightarrow B_{0}=\frac{\mu_{0} I}{4 r} \quad \quad \mathrm{D} \rightarrow \mathrm{II}$
$$ B_{0}=\frac{\mu_{0} I}{2 \pi r}(\pi-1) \quad A \rightarrow \text { III } $$
$B \rightarrow B_{0}=\frac{\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{4 r}+\frac{\mu_{0} I}{4 \pi r}$
$$ B_{0}=\frac{\mu_{0} I}{4 \pi r}(\pi+2) \quad B \rightarrow I $$
$C \rightarrow B_{0}=\frac{\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{4 r}+0$
$B_{0}=\frac{\mu_{0} I}{4 \pi r}(\pi+1) \quad \quad \mathrm{C} \rightarrow \mathrm{IV}$
D. $\rightarrow B_{0}=\frac{\mu_{0} I}{4 r} \quad \quad \mathrm{D} \rightarrow \mathrm{II}$
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