Question
A $13.0 \mathrm{~g}$ wire of length $L=62.0 \mathrm{~cm}$ is suspended by a pair of flexible leads in a uniform magnetic field of magnitude $0.440 \mathrm{~T}$ (Fig. $28-41$ ). What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?
Step 1
Step 1: We know that the magnetic force on a wire carrying current in a magnetic field is given by the formula $F_B = I \cdot B \cdot L$, where $I$ is the current, $B$ is the magnetic field, and $L$ is the length of the wire. Show more…
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A 13.0 g wire of length L = 62.0 cm is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.440 T (figure on the right). What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?
A 13.0 g wire of length $L=62.0 \mathrm{cm}$ is suspended by a pair of flexible leads in a uniform mag- nctic ficld of magnitude 0.440 $\mathrm{T}$ Fig. $28-41$ . What are the (a) magni- tude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?
A 13.0 g wire of length L = 62.0 cm is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.440 T (see figure at at right). What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads? Assume the leads do not act as a spring pulling the wire upwards. Acceleration due to gravity is 9.8 m s⁻¹.
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