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KCET · Physics · Electrostatics

A straight wire of length \( 50 \mathrm{~cm} \) carrying a current of \( 2.5 \mathrm{~A} \) is suspended in mid-air by a uniform magnetic field of \( 0.5 \mathrm{~T} \) (as shown in figure). The mass of the wire is \( \left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right) \)

  1. A \( 62.5 \mathrm{gm} \)
  2. B \( 250 \mathrm{gm} \)
  3. C \( 125 \mathrm{gm} \)
  4. D \( 100 \mathrm{gm} \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( 62.5 \mathrm{gm} \)

Step-by-step Solution

Detailed explanation

Given, the wire is suspended in mid air by uniform magnetic field. Now,
\(\Rightarrow F=m g \)
\(\Rightarrow I B l=m g \)
\(\Rightarrow m=\frac{I B l}{g} \)
\(\Rightarrow m=\frac{2.5 \times 0.5 \times\left(50 \times 10^{-2}\right)}{10}=0.625 \mathrm{~kg}=62.5 \mathrm{~g}\)
Thus, mass of wire is \( 62.5 \mathrm{~g} \).