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,
\[ I \approx 0.27 \, {A} \]
