Question

A block of mass 1 g is equilibrium with the help of a current carrying square loop which is partially lying in constant magnetic field (B) as shown. Resistance of the loop is 10 Ω. Find the voltage (V) (in volts) of the battery in the loop.

Answer 1

Correct Answer: 10

Electromagnetic Force and Current: 

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

\[ F_m = i L B \]

Equating with the gravitational force \( F_m = mg \), we get:

\[ i L B = mg \]

Solving for \(i\):

\[ i = \frac{mg}{L B} \]

Substitute the given values:

• \(m = 1 \times 10^{-3} \ \text{kg}\)
• \(g = 10 \ \text{m/s}^2\)
• \(L = 0.1 \ \text{m}\)
• \(B = 0.1 \ \text{T}\)

 

\[ i = \frac{(1 \times 10^{-3})(10)}{(0.1)(0.1)} \]

\[ i = \frac{1 \times 10^{-2}}{0.01} = 1 \ \text{A} \]

Voltage Across the Loop:

The resistance of the loop is given as \( R = 10 \ \Omega \). Using Ohm's Law:

\[ V = i R \]

Substitute \(i = 1 \ \text{A}\) and \(R = 10 \ \Omega\):

\[ V = (1)(10) = 10 \ \text{V} \]

Final Answer:

\(V = 10 \ \text{V}\)