JEE Main & Advanced
Sample Paper
JEE Main - Mock Test - 27
question_answer
A straight horizontal conducting rod of length 50 cm and mass 50 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero? (Take\[g=10\,\text{m}\,{{\text{s}}^{-2}}\])
A)0.05 T
B)0.1 T
C)0.2 T
D)0.5 T
Correct Answer:
C
Solution :
[c] : Here, \[m=50g=50\times {{10}^{-3}}kg;I=5.0A\] Tension in the wires is zero if the force on the rod due to magnetic field is equal and opposite to the weight of the rod. i.e.,\[mg=BIl\Rightarrow B=\frac{mg}{Il}\] Substituting the given values, we get \[B=\frac{50\times {{10}^{-3}}\times 10}{5\times 50\times {{10}^{-2}}}=0.2T\]