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GUJCET · Physics · Moving Charges and Magnetism
The horizontal component of the earth's magnetic field at a certain place is \(3.0 \times 10^{-5} T\) and the direction of the field is from the geographic south to the geographic north. Every long straight conductor is carrying a steady current of 2 A . What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is east to west.
- A \(9 \times 10^{-5} N / m\)
- B \(6 \times 10^{-5} N / m\)
- C \(3 \times 10^{-5} N / m\)
- D \(2 \times 10^{-5} N / m\)
Answer & Solution
Correct Answer
(B) \(6 \times 10^{-5} N / m\)
Step-by-step Solution
Detailed explanation
B
\(F =i l B\)
force per unit length
\(\begin{array}{l}\frac{ F }{l}=i B \\ \frac{ F }{l}=2 \times 3 \times 10^{-5} \\ \frac{F}{l}=6 \times 10^{-5} N / m \end{array}\)
\(F =i l B\)
force per unit length
\(\begin{array}{l}\frac{ F }{l}=i B \\ \frac{ F }{l}=2 \times 3 \times 10^{-5} \\ \frac{F}{l}=6 \times 10^{-5} N / m \end{array}\)
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