JEE Mains-Numerical

1. Magnetic Field at a Point Midway Between Two Parallel Wires

Shortcut: Use the formula for the magnetic field due to a long straight wire:

$$B = \frac{\mu_0 I}{2\pi d}$$

where:

• $$B$$ is the magnetic field in teslas (T)
• $$\mu_0$$ is the permeability of free space ($$4\pi\times10^{–7}\text{ T}\cdot \text{m/A}$$)
• $$I$$ is the current in amperes (A)
• $$d$$ is the distance from the wire in meters (m)

Solution:

In this problem, we have two wires carrying currents of 5 A and 10 A, respectively. The distance between the wires is 0.1 m. To find the magnetic field at a point midway between the two wires, we can use the formula above.

$$B = \frac{\mu_0}{2\pi d}\left(I_1 + I_2\right)$$

$$B = \frac{(4\pi\times10^{–7}\text{ T}\cdot \text{m/A})}{2\pi (0.1\text{ m})}(5\text{ A} + 10\text{ A})$$

$$B = 2\times10^{–6}\text{ T}$$

Therefore, the magnetic field at a point midway between the two wires is 2 × 10^–6 T.

2. Magnetic Field at the Centre of a Circular Coil

Shortcut: Use Ampere’s law to calculate the magnetic field at a point inside of a loop of current.

$$B =\frac{\mu_0 I}{2R}$$

where,

• $$B$$ is the magnetic field
• $$\mu_0$$ is vacuum permeability
• $$I$$ is the current
• $$R$$ is the radius of the coil

Solution:

Given $$I = 2 \text{A}$$ $$N= 100, R = 0.2 \text{m}$$

$$B =\frac{\mu_0NI}{2R}$$ $$B =\frac{(4\pi\times10^{–7}\text{ T}\cdot \text{m/A}) (100)( 2 \text{A})}{2( 0.2\text{ m)}}$$ $$B = 6.28\times 10^{–4} \text{T}$$

3. Electric Field between the Plates of a Parallel-Plate Capacitor

Shortcut: Use the formula for the electric field between the plates of a parallel-plate capacitor:

$$E = \frac{V}{d}$$

where:

• $$E$$ is the electric field in volts per meter (V/m)
• $$V$$ is the potential difference in volts (V)
• $$d$$ is the distance between the plates in meters (m)

Solution:

In this problem, we have a parallel-plate capacitor with a plate separation of 0.1 mm (0.0001 m) and a potential difference of 100 V. To find the electric field between the plates, we can use the formula above.

$$E = \frac{V}{d} = \frac{100\text{ V}}{0.0001\text{ m}} = 1\times10^6\text{ V/m}$$

Therefore, the electric field between the plates of the capacitor is 1 × 10^6 V/m.

4. Speed of an Electromagnetic Wave

Shortcut: Use the formula for the speed of an electromagnetic wave:

$$v = f\lambda$$

where:

• $$v$$ is the speed of the wave in meters per second (m/s)
• $$f$$ is the frequency of the wave in hertz (Hz)
• $$\lambda$$ is the wavelength of the wave in meters (m)

Solution:

In this problem, we have an electromagnetic wave with a frequency of 1 MHz (1 × 10^6 Hz) and a wavelength of 300 m. To find the speed of the wave, we can use the formula above.

$$v = f\lambda = (1\times10^6\text{ Hz})(300\text{ m}) = 3\times10^8\text{ m/s}$$

Therefore, the speed of the electromagnetic wave is 3 × 10^8 m/s.

5. Amplitude of the Electric Field of a Radio Transmitter

Shortcut: Use the formula for the amplitude of the electric field of a radio transmitter:

$$E_0 = \frac{\sqrt{P}}{\pi d}$$

where:

• $$E_0$$ is the amplitude of the electric field in volts per meter (V/m)
• $$P$$ is the power of the transmitter in watts (W)
• $$d$$ is the distance from the transmitter in meters (m)

Solution:

In this problem, we have a radio transmitter with a power of 100 kW (100,000 W) and a distance of 1 km (1000 m) from the transmitter. To find the amplitude of the electric field, we can use the formula above.

$$E_0 = \frac{\sqrt{P}}{\pi d} = \frac{\sqrt{100,000\text{ W}}}{\pi (1000\text{ m})} = 28.28\text{ V/m}$$

Therefore, the amplitude of the electric field at a distance of 1 km from the transmitter is 28.28 V/m.

CBSE Board Exams- Numericals

1. Magnetic Field at a Point Midway Between Two Parallel Wires

Shortcut: Use the formula for the magnetic field due to a long straight wire:

$$B = \frac{\mu_0 I}{2\pi d}$$

where:

• $$B$$ is the magnetic field in teslas (T)
• $$\mu_0$$ is the permeability of free space ($$4\pi\times10^{–7}\text{ T}\cdot \text{m/A}$$)
• $$I$$ is the current in amperes (A)
• $$d$$ is the distance from the wire in meters (m)

Solution:

In this problem, we have two wires carrying currents of 2 A and 3 A, respectively. The distance between the wires is 0.2 m. To find the magnetic field at a point midway between the two wires, we can use the formula above.

$$B = \frac{\mu_0}{2\pi d}\left(I_1 + I_2\right)$$

$$B = \frac{(4\pi\times10^{–7}\text{ T}\cdot \text{m/A})}{2\pi (0.2\text{ m})}(2\text{ A} + 3\text{ A})$$

$$B = 5\times10^{–6}\text{ T}$$

Therefore, the magnetic field at a point midway between the two wires is 5 × 10^–6 T.

2. Magnetic Field at the Centre of a Circular Coil

Shortcut: Use Ampere’s law to calculate the magnetic field at a point inside of a loop of current.

$$B =\frac{\mu_0 I}{2R}$$

where,

• $$B$$ is the magnetic field
• $$\mu_0$$ is vacuum permeability
• $$I$$ is the current
• $$R$$ is the radius of the coil

Solution:

Given $$I = 1 \text{A}$$ $$N= 50, R = 0.2 \text{m}$$

$$B =\frac{\mu_0NI}{2R}$$ $$B =\frac{(4\pi\times10^{–7}\text{ T}\cdot \text{m/A}) (50)( 1\text{A})}{2( 0.2\text{ m)}}$$ $$B = 3.14\times 10^{–4} \text{T}$$

3. Electric Field between the Plates of a Parallel-Plate Capacitor

Shortcut: Use the formula for the electric field between the plates of a parallel-plate capacitor:

$$E = \frac{V}{d}$$

where:

• $$E$$ is the electric field in volts per meter (V/m)
• $$V$$ is the potential difference in volts (V)
• $$d$$ is the distance between the plates in meters (m)

Solution:

In this problem, we have a parallel-plate capacitor with a plate separation of 0.2 mm (0.0002 m) and a potential difference of 50 V. To find the electric field between the plates, we can use the formula above.

$$E = \frac{V}{d} = \frac{50\text{ V}}{0.0002\text{ m}} = 2.5\times10^5\text{ V/m}$$

Therefore, the electric field between the plates of the capacitor