A 1 m long metal rod XY completes the circuit as shown in f igure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is 5ω, the force needed to move the rod in direction, as indicated, with a constant speed of 4 m/s will be ____10−3 N
Show Hint
Formulas for motional emf \((E = Blv)\) and the force on a current carrying conductor in a magnetic field \((F = BIl)\). Make sure your units are con sistent
Correct Answer: 18
Solution and Explanation
Step 1: Write the Force Equation
The force on a conductor in a magnetic field is given by:
\( F = I \ell B \)
Where:
- \( I \): Current
- \( \ell \): Length of the conductor in the magnetic field
- \( B \): Magnetic field strength
Step 2: Substitute \( I \) in Terms of \( e, R, \) and \( v \)
The current \( I \) can be expressed as:
\( I = \frac{e}{R} \)
Substitute \( I \) into the force equation:
\[ F = Bv \ell B \cdot \frac{\ell}{R} \]
Simplify:
\[ F = \frac{B^2 \ell^2 v}{R} \]
Step 3: Substitute Known Values
Given:
- \( B = 15 \, \text{T} \)
- \( \ell = 1 \, \text{m} \)
- \( v = 4 \, \text{m/s} \)
- \( R = 5 \, \Omega \)
Substitute these values into the equation:
\[ F = \frac{(15)^2 \cdot (1)^2 \cdot 4}{5} \]
Simplify:
\[ F = \frac{225 \cdot 4}{5} = 180 \, \text{N} \]
Final Answer:
The magnetic force is \( F = 18 \, \text{N}. \)
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