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Q.(a) Write the vector form of the Biot-Savart law.

(b) Two insulated long straight wires, each carrying 2.02.0 A current, are kept along the xx′xx' and yy′yy' axes as shown in the figure. Find the magnitude and direction of the resultant magnetic field at point P (4 m,5 m)(4\,\text{m}, 5\,\text{m}).
Figure — 55/4/1 Q23
Figure
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The Biot-Savart law gives the magnetic field due to a current element. For two perpendicular long straight wires, the net field at a point is the vector sum of the fields from each wire, found using B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}. At P(4,5), the fields are 8.0×10−8 T8.0\times10^{-8}\,\text{T} (out of page) and 1.0×10−7 T1.0\times10^{-7}\,\text{T} (into page), giving a resultant of 2.0×10−8 T2.0\times10^{-8}\,\text{T} into the page.

The Concept

The Biot-Savart law is the fundamental relation that tells us how a current-carrying conductor produces a magnetic field. For a long straight wire, symmetry simplifies this to a neat result: the field at a perpendicular distance rr from the wire has magnitude B=μ0I2πrB = \frac{\mu_0 I}{2\pi r} and circles the wire according to the right-hand rule.

When two wires are present, the principle of superposition applies — the net magnetic field is simply the vector sum of the individual fields. The key is to treat each wire's contribution independently, then add them as vectors.

Step-by-Step Solution

Part (a): Vector form of Biot-Savart law

The Biot-Savart law in vector form states that the magnetic field dB⃗d\vec{B} produced by a current element I dl⃗I\,d\vec{l} at a point located by the position vector r⃗\vec{r} (from the element to the point) is:

dB⃗=μ04πI dl⃗×r^r2d\vec{B} = \frac{\mu_0}{4\pi} \frac{I\,d\vec{l} \times \hat{r}}{r^2}

where r^=r⃗r\hat{r} = \frac{\vec{r}}{r} is the unit vector from the current element to the point, and μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7}\,\text{T m/A} is the permeability of free space.

dB⃗=μ04πI dl⃗×r^r2d\vec{B} = \frac{\mu_0}{4\pi} \frac{I\,d\vec{l} \times \hat{r}}{r^2}

Part (b): Magnetic field at point P

1. Identify the geometry.

Figure — 55/4/1 Q23
Figure — 55/4/1 Q23

Wire 1 lies along the xx′xx'-axis (the x-axis). Wire 2 lies along the yy′yy'-axis (the y-axis). Point P has coordinates (4 m,5 m)(4\,\text{m}, 5\,\text{m}).

The perpendicular distance from P to the x-axis wire is simply the y-coordinate: r1=5 mr_1 = 5\,\text{m}.

The perpendicular distance from P to the y-axis wire is the x-coordinate: r2=4 mr_2 = 4\,\text{m}.

2. Determine the direction of each field using the right-hand rule.

For a long straight wire, point your thumb in the direction of the current, and your fingers curl in the direction of the magnetic field lines.

  • Wire along x-axis: The current is along the xx′xx' direction. At point P (which has positive y), the field circles the wire. Using the right-hand rule, the field at P points out of the page (along +k^+\hat{k}).
  • Wire along y-axis: The current is along the yy′yy' direction. At point P (which has positive x), the field circles this wire. Using the right-hand rule, the field at P points into the page (along −k^-\hat{k}).
Watch out

A common mistake is to get the direction wrong for the y-axis wire. Draw it: thumb pointing up (along +y+y), and at a point to the right (positive x), your fingers curl into the page. Always sketch the geometry. …

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