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A rectangular conducting loop consists of two wires on two opposite sides of length 1 joined together by rods of length d. The wires are each of the same material but with cross sections differing by a factor of 2. The thicker wire has a resistance R and the rods are of low resistance, which in turn are connected to a constant voltage source V_(0). The loop is placed in uniform a magnetic field vecB at 45^(@) to its plane. Find the torque exerted by the magnetic field on the loop about an axis through the centres of rods. |
Answer» Solution :1.   2. Resistance of thick wire be R and THIN wire be 2R. Cross section is different for both wire from factor q. MASS and length of both are same. 3. Torque and force on first wire is as shown below, `F_(1)=I_(1)lB=V_(0)/(2R)lB` `thereforetau_(1)=(F_(1))(d/(2sqrt2))=(V_(0)lB(d))/(2R2sqrt2)` 4. Torque and force on another wire can be calculated as, `F_(2)=I_(2)lB=V_(0)/(2R)lB` `thereforetau_(2)=F_(2)(d/(2sqrt2))` `thereforetau_(2)=(V_(0)lB)/(2R)(1/(2sqrt2))` 5. According to Ohm.s law, V = IR 6. Current `I_(1)=V_(1)/R_(1)=V_(0)/R""(becauseV_(1)=V_(2)=V_(0))` 7. Current `I_(2)=V_(2)/R_(2)=V_(0)/(2R)` 8. Force on first wire, `F_(1)=I_(1)lB(1/sqrt2)` = `V_(0)/R(lB)/sqrt2` `therefore` Torque on wire, `tau_(1)=F_(1)(d)` `tau_(1)=(V_(0)lB)/(sqrt2R)(d)` 9. Similarly, force and torque can be calculated for second wire are, `F_(2)=I_(2)lBsin45^(@)` = `V_(0)/(2R)(lB)/sqrt2` `thereforetau_(2)=F_(2)(d)` = `V_(0)/(2R)(lB)/sqrt2(d)` `tau_(2)=(V_(0)lB)/(2sqrt(2)R)(d)` 10. RESULTANT torque, `tau=tau_(1)-tau_(2)=(V_(0)lB)/(sqrt2R)d/2-(V_(0)lBd)/(2sqrt2Rxx2)` `thereforetau=1/(4sqrt2)(V_(0)lB)/R(d)=1/(4sqrt2)(V_(0)BA)/R` where ld = area A |
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