A wire of 60 cm length and mass 10 g is suspended by a pair of flexible leads in a magnetic field of 0.60 T as shown in the figure. The magnitude of the current required to remove the tension in the supporting leads is:
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To balance the gravitational force with magnetic force in a wire suspended in a magnetic field, ensure the current direction is such that the Lorentz force opposes gravity.
Step 1: Calculate the magnetic force needed to balance gravity.
The gravitational force \( F_g \) is given by:
\[ F_g = mg \]
where \( m = 0.01 \, {kg} \) (mass of the wire), \( g = 9.81 \, {m/s}^2 \) (acceleration due to gravity).
\[ F_g = 0.01 \, {kg} \times 9.81 \, {m/s}^2 = 0.0981 \, {N} \]
Step 2: Apply the formula for magnetic force.
The magnetic force \( F_m \) that balances the weight is given by:
\[ F_m = ILB \]
where \( I \) is the current, \( L = 0.6 \, {m} \) (length of the wire), and \( B = 0.6 \, {T} \) (magnetic field strength).
Setting \( F_m = F_g \),
\[ 0.6 \times I \times 0.6 = 0.0981 \]
\[ I = \frac{0.0981}{0.36} \]
\[ I = 0.2725 \, {A} \]
Rounded to two significant figures,
\[ I \approx 0.27 \, {A} \]
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Approach Solution -2
To solve this problem, we need to calculate the current required to remove the tension in the supporting leads of the wire in a magnetic field.
1. Given Data:
We are given:
The length of the wire \( L = 60 \, \text{cm} = 0.6 \, \text{m} \)
The mass of the wire \( m = 10 \, \text{g} = 0.01 \, \text{kg} \)
The magnetic field strength \( B = 0.60 \, \text{T} \)
Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)
2. Forces Acting on the Wire:
The tension in the supporting leads must balance the force due to gravity and the magnetic force on the current-carrying wire. The force due to gravity is:
\[
F_{\text{gravity}} = m \cdot g = 0.01 \cdot 9.8 = 0.098 \, \text{N}
\]
The magnetic force on the wire is given by:
\[
F_{\text{magnetic}} = B \cdot I \cdot L
\]
where \( I \) is the current flowing through the wire, \( B \) is the magnetic field strength, and \( L \) is the length of the wire in the magnetic field.
For equilibrium, the magnetic force must balance the gravitational force, so:
\[
B \cdot I \cdot L = F_{\text{gravity}} = 0.098
\]
Solving for \( I \):
\[
I = \frac{0.098}{B \cdot L} = \frac{0.098}{0.60 \cdot 0.6} = \frac{0.098}{0.36} = 0.27 \, \text{A}
\]
