JEE Main & JEE Advanced:

1. Numerical on Lenz’s law:

• Calculate the magnetic force acting on the rod: $$F_B=BIL$$ $$F_B=(0.5T)(1m)(100g)(9.8m/s^2)=4.9N$$
• Use the right-hand rule to determine the direction of the induced current. The thumb points in the direction of motion, the fingers point in the direction of the magnetic field, and the middle finger points in the direction of the induced current. In this case, the induced current flows from the left rail to the right rail.

2. Numerical on motional emf:

• Calculate the area of the loop: $$A=l\times w=(10cm)(20cm)=200c{m}^2$$ $$A=0.02m^2$$
• Calculate the magnetic flux through the loop: $$\mathrm{\Phi }=BAcos\theta$$ $$\mathrm{\Phi }=(0.2T)(0.02m^2)cos0\text{degree}$$ $$\mathrm{\Phi }=0.004Wb$$
• Calculate the induced emf: $$emf=-\frac{d\mathrm{\Phi }}{dt}$$ $$emf=-\frac{(0.004Wb-0Wb)}{1s-0s}$$ $$emf=4mV$$

3. Numerical on Faraday’s law:

• Calculate the change in magnetic flux through the solenoid: $$\mathrm{\Delta }\mathrm{\Phi }=NI(\mathrm{\Delta }t)$$ $$\mathrm{\Delta }\mathrm{\Phi }=(1000turns)(5A)(0.1s)$$ $$\mathrm{\Delta }\mathrm{\Phi }=50Wb$$
• Calculate the induced emf: $$emf=-\frac{d\mathrm{\Phi }}{dt}$$ $$emf=-\frac{(50Wb-0Wb)}{(0.1s)-0s}$$ $$emf=500V$$

4. Numerical on transformer action:

• Calculate the turns ratio of the transformer: $$\frac{N_p}{N_s}=\frac{V_p}{V_s}$$ $$\frac{100turns}{2000turns}=\frac{12V}{V_s}$$ $$V_s=\frac{(12V)(2000turns)}{100turns}$$ $$V_s=240V$$

5. Numerical on self-inductance:

• Calculate the induced emf: $$emf=L\frac{\mathrm{\Delta }I}{\mathrm{\Delta }t}$$ $$2V=L\left(\frac{1A-0A}{0.1s-0s}\right)$$ $$2V=10L$$
• Calculate the self-inductance: $$L=0.2H$$

CBSE Board Exam:

1. Numerical on Lenz’s law:

• Calculate the magnetic force acting on the rod: $$F_B=BIL$$ $$F_B=(0.5T)(0.2m)(5m/s)=0.5N$$
• Use the right-hand rule to determine the direction of the induced current. The thumb points in the direction of motion, the fingers point in the direction of the magnetic field, and the middle finger points in the direction of the induced current. In this case, the induced current flows from the bottom of the rod to the top of the rod.

2. Numerical on motional emf:

• Calculate the area of the loop: $$A=l\times w=(5cm)(10cm)=50c{m}^2$$ $$A=0.005m^2$$
• Calculate the magnetic flux through the loop: $$\mathrm{\Phi }=BAcos\theta$$ $$\mathrm{\Phi }=(0.1T)(0.005m^2)cos0\text{degree}$$ $$\mathrm{\Phi }=0.0005Wb$$
• Calculate the induced emf: $$emf=-\frac{d\mathrm{\Phi }}{dt}$$ $$emf=-\frac{(0.0005Wb-0Wb)}{1s-0s}$$ $$emf=0.5mV$$

3. Numerical on Faraday’s law:

• Calculate the change in magnetic flux through the solenoid: $$\mathrm{\Delta }\mathrm{\Phi }=NI(\mathrm{\Delta }t)$$ $$\mathrm{\Delta }\mathrm{\Phi }=(500turns)(2A)(0.05s)$$ $$\mathrm{\Delta }\mathrm{\Phi }=50Wb$$
• Calculate the induced emf: $$emf=-\frac{d\mathrm{\Phi }}{dt}$$ $$emf=-\frac{(50Wb-0Wb)}{0.05s-0s}$$ $$emf=1000V$$

4. Numerical on transformer action:

• Calculate the turns ratio of the transformer: $$\frac{N_p}{N_s}=\frac{V_p}{V_s}$$ $$\frac{500turns}{100turns}=\frac{220V}{V_s}$$ $$V_s=\frac{(220V)(100turns)}{500turns}$$ $$V_s=44V$$

5. Numerical on self-inductance:

• Calculate the induced emf: $$emf=L\frac{\mathrm{\Delta }I}{\mathrm{\Delta }t}$$ $$1V=L\left(\frac{0.5A-0A}{0.05s-0s}\right)$$ $$1V=10L$$
• Calculate the self-inductance: $$L=0.1H$$