MHT CET · Physics · Magnetic Properties of Matter
A horizontal wire of mass ' \(\mathrm{m}\) ', length ' \(l\) ' and resistance ' \(R\) ' is sliding on the vertical rails on which uniform 'magnetic field ' \(B\) ' is directed perpendicular. The terminal speed of the wire as it falls under the force of gravity is ( \(\mathrm{g}=\) acceleration due to gravity)
- A \(\frac{\mathrm{mgl}}{\mathrm{BR}}\)
- B \(\frac{\mathrm{B}^2 l^2}{\mathrm{mgR}}\)
- C \(\frac{\mathrm{mgR}}{\mathrm{B} l}\)
- D \(\frac{\mathrm{mgR}}{\mathrm{B}^2 l^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{mgR}}{\mathrm{B} l}\)
Step-by-step Solution
Detailed explanation
Net force on the wire becomes zero when it attains terminal velocity.
\(\therefore \quad\) Force due to magnetic field = gravitational force
\(\therefore \quad \mathrm{iB} l=\mathrm{mg}\)
\(\therefore \quad \frac{\mathrm{e}}{\mathrm{R}} \mathrm{B} l=\mathrm{mg} \quad \ldots\left(\because \mathrm{i}=\frac{\mathrm{e}}{\mathrm{R}}\right)\)
\(\therefore \quad \frac{\mathrm{Bv} l}{\mathrm{R}} \mathrm{B} l=\mathrm{mg} \quad \ldots(\because \mathrm{e}=\mathrm{Bv} l)\)
\(\therefore \quad \mathrm{v}=\frac{\mathrm{mgR}}{\mathrm{B}^2 l^2}\)
\(\therefore \quad\) Force due to magnetic field = gravitational force
\(\therefore \quad \mathrm{iB} l=\mathrm{mg}\)
\(\therefore \quad \frac{\mathrm{e}}{\mathrm{R}} \mathrm{B} l=\mathrm{mg} \quad \ldots\left(\because \mathrm{i}=\frac{\mathrm{e}}{\mathrm{R}}\right)\)
\(\therefore \quad \frac{\mathrm{Bv} l}{\mathrm{R}} \mathrm{B} l=\mathrm{mg} \quad \ldots(\because \mathrm{e}=\mathrm{Bv} l)\)
\(\therefore \quad \mathrm{v}=\frac{\mathrm{mgR}}{\mathrm{B}^2 l^2}\)
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