JEE Advanced Numerical Examples

1. To find the magnitude of the magnetic field at a point (2, 0, 0) due to a long straight wire carrying a current of 10 A, we use the Biot-Savart law:

$$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\int \frac{I\overrightarrow{dl}\times \hat{r}}{r^2}$$ Where

• ${\mu }_0$ is the permeability of free space
• (\overrightarrow{dl}) is the differential length of the wire
• (\hat{r}) is the unit vector from the current element to the point of observation

Since the wire is long and straight, we can assume that r = 2 cm for all points on the wire. Also, since the observation point is at right angles to the wire, the cross product (\overrightarrow{dl} \times \hat{r}) is parallel to the z-axis. Therefore:

$$|\overrightarrow{B}|=\frac{{\mu }_{0}I}{4\pi r}\int dl$$ $$|\overrightarrow{B}|=\frac{{\mu }_{0}I}{4\pi r}2\pi r$$ $$|\overrightarrow{B}|=\frac{{\mu }_{0}I}{2}$$ Substituting the values, we get: $$|\overrightarrow{B}|=\frac{(4\times {10}^{-7}Tm/A)(10A)}{2}$$ $$|\overrightarrow{B}|=2\times {10}^{-6}T$$

2. To find the magnitude of the magnetic field at the center of a circular coil of radius 5 cm carrying a current of 2 A, we use the formula for the magnetic field of a circular loop:

$$B=\frac{{\mu }_{0}NI}{2R}$$

Where

• (\mu_0) is the permeability of free space
• (N) is the number of turns
• (I) is the current
• (R) is the radius of the coil

Substituting the values, we get: $$B=\frac{(4\times {10}^{-7}Tm/A)(100)(2A)}{2(0.05m)}$$ $$B=4\times {10}^{-4}T$$

3. To find the magnitude of the magnetic field inside a solenoid of length 20 cm, having 1000 turns and carrying a current of 5 A, we use the formula:

$$B={\mu }_{0}nI$$

Where

• (\mu_0) is the permeability of free space
• (n) is the number of turns per unit length
• (I) is the current

In this case, the length of the solenoid is 20 cm which is 0.2 m. Therefore, the number of turns per unit length is: $$n=\frac{1000turns}{0.2m}$$ $$n=5000turns/m$$ Substituting the values into the formula, we get: $$B=(4\times {10}^{-7}Tm/A)(5000turns/m)(5A)$$ $$B=0.1T$$

4. To find the magnitude of the magnetic field at a point 5 cm away from the center of a bar magnet of length 10 cm and magnetic moment 0.5 A-m2, we use the formula for the magnetic field on axial line of bar magnet:

$$B=\frac{2{\mu }_{0}m}{4\pi r^3}$$

Where

• (\mu_0) is the permeability of free space
• (m) is the magnetic moment
• (r) is the distance from the center of the magnet

Substituting the values, we get: $$B=\frac{2(4\times {10}^{-7}Tm/A)(0.5A{m}^2)}{4\pi (0.05m{)}^3}$$ $$B=3.18\times {10}^{-4}T$$

5. To find the magnitude of the magnetic force acting on a moving charge of 2 C moving with a velocity of 10 m/s in a magnetic field of 0.5 T perpendicular to the velocity, we use the formula:

$$F=qvB\sin \theta$$

Where

• (q) is the charge
• (v) is the velocity
• (B) is the magnetic field strength
• (\theta) is the angle between (v) and (B)

In this case, (\theta) is 90 degrees, so (\sin\theta=1). Therefore,

$$F=(2C)(10m/s)(0.5T)(1)$$ $$F=10N$$

CBSE Board Exam Numerical Examples

1. The magnetic field at a point (1, 0, 0) due to a long straight wire carrying a current of 5 A can be calculated using the Biot-Savart law:

$$B=\frac{{\mu }_{0}I}{4\pi r}$$

Where

• (\mu_0) is the permeability of free space ((4\times10^{-7}Tm/A))
• (I) is the current (5A)
• (r) is the distance from the point to the wire (1 m)

Substituting these values, we get: $$B=\frac{(4\times {10}^{-7}Tm/A)(5A)}{4\pi (1m)}$$ $$B=1\times {10}^{-6}T$$ So the magnetic field at the point (1, 0, 0) is (1\times10^{-6} T).

2. The magnetic field at the center of a circular coil of radius 2 cm carrying a current of 1 A can be calculated using the formula: $$B=\frac{{\mu }_{0}NI}{2R}$$ Where

• (\mu_0) is the permeability of free space ((4\times10^{-7}Tm/A))
• (N) is the number of turns (50)
• (I) is the current (1 A)
• (R) is the radius of the coil (0.02 m)

Substituting these values, we get: $$B=\frac{(4\times {10}^{-7}Tm/A)(50)(1A)}{2(0.02m)}$$ $$B=0.5\times {10}^{-3}T$$ Therefore, the magnitude of the magnetic field produced at the center of the circular coil is (5 \times 10^{-4} T).

3. To calculate the magnitude of the magnetic field inside a solenoid of length 10 cm having 500 turns and carrying a current of 2 A, we can use the formula:

$$B={\mu }_{0}nI$$

Where:

• (\mu_0) is the permeability of free space ((4\pi × 10^{−7} Tm/A))
• (n) is the number of turns per unit length ((n=N/L))
• (I) is the current (2 A)

First, calculate the number of turns per unit length (n):

$$n=\frac{500turns}{0.1m}=5000turns/m$$

Then, substitute the values into the formula: $$B=(4\pi \times {10}^{-7}Tm/A)(5000turns/m)(2A)$$ $$B=4.0\times {10}^{-3}T$$

Therefore, the magnitude of the magnetic field inside the solenoid is (4.0 × 10^{−3} T).

4. To determine the magnitude of the magnetic field at a point 2.5 cm away from the center of a bar magnet of length 5 cm with a magnetic moment of 0.25 A-m2 along its axis, we can use the formula for the magnetic field on the axial line of a bar magnet:

$$B=\frac{2{\mu }_{0}m}{4\pi x^3}$$

Where:

• (B) is the magnetic field strength
• (\mu_0) is the permeability of free space ((4\pi \times 10^{-7} T m/A))
• (m) is the magnetic moment of the bar magnet (0.25 A-m2)
• (x) is the distance from the center of the magnet to the observation point (0.025 m)

By substituting the numerical values, we can get the magnetic field strength: $$B=\frac{2(4\pi \times {10}^{-7}Tm/A)(0.25A{m}^2)}{4\pi (0.025m{)}^3}$$ $$B=1.6\times {10}^{-5}T$$

Hence, the magnitude of the magnetic field at the specified point is (1.6\times10^{-5} T).

5. To calculate the magnitude of the magnetic force acting on a moving charge of 1 C moving with a velocity of 5