Class 10 - Unit # 16 : Electromagnetism - Solved Numericals

Electromagnetism

1. A wire carrying $4\text{ A}$ current and having a length of $15\text{ cm}$ between the poles of a magnet is kept at an angle of $30^\circ$ to a uniform field of $0.8\text{ T}$. Find the force acting on the wire.

Data:

• Current ($I$) = $4\text{ A}$
• Length of wire ($L$) = $15\text{ cm} = 0.15\text{ m}$
• Magnetic field strength ($B$) = $0.8\text{ T}$
• Angle ($\theta$) = $30^\circ$
• Magnetic force ($F$) = ?

Solution:

The force acting on a current-carrying conductor in a uniform magnetic field is given by:

$$F = I \cdot L \cdot B \cdot \sin(\theta)$$ $$F = 4\text{ A} \times 0.15\text{ m} \times 0.8\text{ T} \times \sin(30^\circ)$$ $$F = 0.48 \times 0.5$$ $$F = 0.24\text{ N}$$

RESULT: The magnetic force acting on the wire is $0.24\text{ N}$.

2. A square loop of wire of side $2.0\text{ cm}$ carries $2.0\text{ A}$ of current. A uniform magnetic field of magnitude $0.7\text{ T}$ makes an angle of $30^\circ$ with the plane of the loop. What is the magnitude of torque on the loop?

Data:

• Number of turns ($N$) = $1$
• Side of the square loop ($s$) = $2.0\text{ cm} = 0.02\text{ m}$
• Area of the loop ($A$) = $s^2 = (0.02\text{ m})^2 = 4 \times 10^{-4}\text{ m}^2$
• Current ($I$) = $2.0\text{ A}$
• Magnetic field magnitude ($B$) = $0.7\text{ T}$
• Angle with the plane of the loop ($\alpha$) = $30^\circ$
• Torque ($\tau$) = ?

Solution:

The torque acting on a current-carrying coil when the angle is specified with the plane of the loop is given by:

$$\tau = N \cdot I \cdot A \cdot B \cdot \cos(\alpha)$$ $$\tau = 1 \times 2.0\text{ A} \times (4 \times 10^{-4}\text{ m}^2) \times 0.7\text{ T} \times \cos(30^\circ)$$ $$\tau = 5.6 \times 10^{-4} \times 0.8660$$ $$\tau \approx 4.85 \times 10^{-4}\text{ N}\cdot\text{m}$$

RESULT: The magnitude of the torque acting on the square loop is $4.85 \times 10^{-4}\text{ N}\cdot\text{m}$.

Electromagnetic Induction

3. A transformer is needed to convert a mains $220\text{ V}$ supply into a $12\text{ V}$ supply. If there are $2200$ turns on the primary coil, then find the number of turns on the secondary coil.

Data:

• Primary voltage ($V_p$) = $220\text{ V}$
• Secondary voltage ($V_s$) = $12\text{ V}$
• Primary turns ($N_p$) = $2200$
• Secondary turns ($N_s$) = ?

Solution:

According to the transformer turns ratio equation:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies N_s = \left(\frac{V_s}{V_p}\right) \cdot N_p$$ $$N_s = \left(\frac{12}{220}\right) \times 2200$$ $$N_s = 12 \times 10 = 120\text{ turns}$$

RESULT: The total number of turns required on the secondary coil is $120\text{ turns}$.

4. A coil surrounds a long solenoid. The current in the solenoid is changing at a rate of $150\text{ A/s}$ and the mutual inductance of the two coils is $5.5 \times 10^{-5}\text{ H}$. Determine the electromotive force (EMF) induced in the surrounding coil.

Data:

• Rate of change of current in primary ($\frac{\Delta I_p}{\Delta t}$) = $150\text{ A/s}$
• Mutual inductance ($M$) = $5.5 \times 10^{-5}\text{ H}$
• Induced electromotive force ($E_s$) = ?

Solution:

According to Faraday's law of mutual induction, the magnitude of induced EMF is given by:

$$E_s = M \cdot \frac{\Delta I_p}{\Delta t}$$ $$E_s = (5.5 \times 10^{-5}\text{ H}) \times 150\text{ A/s}$$ $$E_s = 825 \times 10^{-5} = 8.25 \times 10^{-3}\text{ V}$$

RESULT: The magnitude of the EMF induced in the surrounding secondary coil is $8.25 \times 10^{-3}\text{ V}$ (or $8.25\text{ mV}$).