Shortcut Methods
JEE Advanced Numerical Examples
1. Magnetic field due to a long straight wire:
- Formula: $$B = \frac{\mu_0 I}{2\pi r}$$
- Given values: (I = 10\text{ A}), ((x, y, z) = (3, 4, 0)\text{ m})
- Calculation: $$B = \frac{(4\pi \times 10^{-7}\text{ Tm/A})(10\text{ A})}{2\pi(3\text{ m})}$$
- $$B = 6.67 \times 10^{-7} \text{ T}$$
2. Magnetic field at the center of a circular loop:
- Formula: $$B = \frac{\mu_0 I}{2R}$$
- Given values: (I = 2\text{ A}), (R = 5 \times 10^{-2} \text{ m})
- Calculation: $$B = \frac{(4\pi \times 10^{-7}\text{ Tm/A})(2\text{ A})}{2(5 \times 10^{-2} \text{ m})}$$
- $$B = 2.51 \times 10^{-5} \text{ T}$$
3. Magnetic field inside a solenoid:
- Formula: $$B = \mu_0 nI$$
- Given values: (n = \frac{1000 \text{ turns}}{0.2\text{ m}} = 5000 \text{ turns/m}), (I = 2\text{ A})
- Calculation: $$B = (4\pi \times 10^{-7}\text{ Tm/A})(5000 \text{ turns/m})(2\text{ A})$$
- $$B = 0.126 \text{ T}$$
4. Magnetic force on a moving proton:
- Formula: $$F = qvB\sin\theta$$
- Given values: (q = 1.6 \times 10^{-19}\text{ C}), (v = 10^6 \text{ m/s}), (B = 1\text{ T}), (\theta = 90^\circ)
- Calculation: $$F = (1.6 \times 10^{-19}\text{ C})(10^6 \text{ m/s})(1\text{ T})\sin90^\circ$$
- $$F = 1.6 \times 10^{-13} \text{ N}$$
5. Frequency of revolution of an electron in a magnetic field:
- Formula: $$f = \frac{qvB}{2\pi m}$$
- Given values: (q = -1.6 \times 10^{-19}\text{ C}), (v = 10^6 \text{ m/s}), (B = 0.1\text{ T}), (m = 9.11 \times 10^{-31} \text{ kg})
- Calculation: $$f = \frac{(-1.6 \times 10^{-19}\text{ C})(10^6 \text{ m/s})(0.1\text{ T})}{2\pi (9.11 \times 10^{-31} \text{ kg})}$$
- $$f = 2.81 \times 10^{12} \text{ Hz}$$
CBSE Board Exam Numerical Examples
1. Magnetic field due to a long straight wire:
- Formula: $$B = \frac{\mu_0 I}{2\pi r}$$
- Given values: (I = 5\text{ A}), ((x, y, z) = (2, 3, 0)\text{ m})
- Calculation: $$B = \frac{(4\pi \times 10^{-7}\text{ Tm/A})(5\text{ A})}{2\pi(2\text{ m})}$$
- $$B = 5 \times 10^{-7} \text{ T}$$
2. Magnetic field at the center of a circular loop:
- Formula: $$B = \frac{\mu_0 I}{2R}$$
- Given values: (I = 1\text{ A}), (R = 2 \times 10^{-2} \text{ m})
- Calculation: $$B = \frac{(4\pi \times 10^{-7}\text{ Tm/A})(1\text{ A})}{2(2 \times 10^{-2} \text{ m})}$$
- $$B = 6.28 \times 10^{-6} \text{ T}$$
3. Magnetic field inside a solenoid:
- Formula: $$B = \mu_0 nI$$
- Given values: (n = \frac{500 \text{ turns}}{0.1 \text{ m}} = 5000 \text{ turns/m}), (I = 1\text{ A})
- Calculation: $$B = (4\pi \times 10^{-7}\text{ Tm/A})(5000 \text{ turns/m})(1\text{ A})$$
- $$B = 6.28 \times 10^{-4} \text{ T}$$
4. Magnetic force on a moving proton:
- Formula: $$F = qvB\sin\theta$$
- Given values: (q = 1.6 \times 10^{-19}\text{ C}), (v = 5 \times 10^5 \text{ m/s}), (B = 0.5\text{ T}), (\theta = 90^\circ)
- Calculation: $$F = (1.6 \times 10^{-19}\text{ C})(5 \times 10^5 \text{ m/s})(0.5\text{ T})\sin90^\circ$$
- $$F = 4 \times 10^{-14} \text{ N}$$
5. Frequency of revolution of an electron in a magnetic field:
- Formula: $$f = \frac{qvB}{2\pi m}$$
- Given values: (q = -1.6 \times 10^{-19}\text{ C}), (v = 10^6 \text{ m/s}), (B = 0.05\text{ T}), (m = 9.11 \times 10^{-31} \text{ kg})
- Calculation: $$f = \frac{(-1.6 \times 10^{-19}\text{ C})(10^6 \text{ m/s})(0.05\text{ T})}{2\pi (9.11 \times 10^{-31} \text{ kg})}$$
- $$f = 8.79 \times 10^{11} \text{ Hz}$$