- 1(current)
Q. No.Given below are two statements:
| Statement I: | The magnetic field due to a segment \({d\vec l}\) of a current-carrying wire carrying a current, \(I\) is given by: \({d\vec B}=\dfrac{\mu_0}{4\pi}~I\left({d\vec l}\times\dfrac{\vec r}{r^3}\right ),\) where \(\vec{r}\) is the position vector of the field point with respect to the wire segment. |
| Statement II: | The magnetic field of a current-carrying wire is never parallel to the wire. |
| 1. | Statement I and Statement II are True and Statement I is the correct explanation of Statement II. |
| 2. | Statement I and Statement II are True and Statement I is not the correct explanation of Statement II. |
| 3. | Statement I is True, and Statement II is False. |
| 4. | Statement I is False, and Statement II is True. |
From NCERT
Subtopic: Biot-Savart Law |
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyThe same current \(I\) is flowing in two infinitely long wires in the positive \(x \) and \(y\)-directions. The magnetic field at a point \((0,0,a)\) would be:
| 1. | \( \dfrac{\mu_{0} i}{2 \pi a}(\hat{i}+\hat{j})\) | 2. | \( \dfrac{\mu_{0} i}{2 \pi a}(-\hat{i}+\hat{j})\) |
| 3. | \(\dfrac{\mu_{0} i}{2 \pi a}(-\hat{i}-\hat{j})\) | 4. | \(\dfrac{\mu_{0} i}{2 \pi a}(\hat{i}-\hat{j})\) |
From NCERT
Subtopic: Biot-Savart Law |
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyAs shown in the figure, the equal current \(I\) flows in the two segments; the magnetic field at the centre of the loop due to segment \(ABC\) is \(B_1\) and due to segment \(ADB\) is \(B_2.\) Then:
1. \(B_1 > B_2\)
2. \(B_1 < B_2\)
3. \(B_1=B_2\)
4. \(2B_1=B_2\)
1. \(B_1 > B_2\)
2. \(B_1 < B_2\)
3. \(B_1=B_2\)
4. \(2B_1=B_2\)
From NCERT
Subtopic: Biot-Savart Law |
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyIdentical cells are connected to identical square wire loops as shown in the two diagrams, and the magnetic fields are respectively \(B_1\) and \(B_2\) at the centres.
Then, we can conclude that:
1. \(B_1>0, B_2=0\)
2. \(B_1> B_2>0\)
3. \(B_2> B_1>0\)
4. \(B_1=0, B_2=0\)
Then, we can conclude that:
1. \(B_1>0, B_2=0\)
2. \(B_1> B_2>0\)
3. \(B_2> B_1>0\)
4. \(B_1=0, B_2=0\)
Subtopic: Biot-Savart Law |
To view explanation, please take trial in the course below.
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology- 1(current)
Q. No.
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