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 ($$4\pi \times {10}^{–7}\text{ T}\cdot \text{m/A}$$)
• $$I$$ 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.

$$B=\frac{{\mu }_0}{2\pi d}(I_1+I_2)$$

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

$$B=2\times {10}^{–6}\text{ T}$$

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

$$B=\frac{{\mu }_{0}NI}{2R}$$ $$B=\frac{(4\pi \times {10}^{–7}\text{ T}\cdot \text{m/A})(100)(2\text{A})}{2(0.2\text{ m)}}$$ $$B=6.28\times {10}^{–4}\text{T}$$

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.

$$E=\frac{V}{d}=\frac{100\text{ V}}{0.0001\text{ m}}=1\times {10}^6\text{ V/m}$$

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.

$$v=f\lambda =(1\times {10}^6\text{ Hz})(300\text{ m})=3\times {10}^8\text{ m/s}$$

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 ($$4\pi \times {10}^{–7}\text{ T}\cdot \text{m/A}$$)
• $$I$$ 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.

$$B=\frac{{\mu }_0}{2\pi d}(I_1+I_2)$$

$$B=\frac{(4\pi \times {10}^{–7}\text{ T}\cdot \text{m/A})}{2\pi (0.2\text{ m})}(2\text{ A}+3\text{ A})$$

$$B=5\times {10}^{–6}\text{ T}$$

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

$$B=\frac{{\mu }_{0}NI}{2R}$$ $$B=\frac{(4\pi \times {10}^{–7}\text{ T}\cdot \text{m/A})(50)(1\text{A})}{2(0.2\text{ m)}}$$ $$B=3.14\times {10}^{–4}\text{T}$$

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.

$$E=\frac{V}{d}=\frac{50\text{ V}}{0.0002\text{ m}}=2.5\times {10}^5\text{ V/m}$$

Therefore, the electric field between the plates of the capacitor