Magnetic Effects & Magnetism

JEE Main level, single correct, medium

61 of 276

Question 61 of 276

A long solenoid carrying a current produces a magnetic field BB along its axis. If the current is doubled and the number of turns per cmcm is halved, the new value of magnetic field will be equal to

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Answer: (D) BB

Magnetic field due to long solenoid along its axis is given as,
B=μ0nIB={\mu }_{0}nI, where nn is the number of turns in the solenoid per unit length, II is current in the solenoid and μ0{\mu }_{0} is a constant called the permeability of free space.
⇒B1B2=n1I1n2I2\Rightarrow \dfrac{{B}_{1}}{{B}_{2}}=\dfrac{{n}_{1}{I}_{1}}{{n}_{2}{I}_{2}}
According to the question, n2=n12andI2=2I1{n}_{2}=\dfrac{{n}_{1}}{2}\text{and}{I}_{2}=2{I}_{1}
⇒BB2=n1(n12)×I12I1⇒B2=B\Rightarrow \dfrac{B}{{B}_{2}}=\dfrac{{n}_{1}}{\left(\dfrac{{n}_{1}}{2}\right)}\times \dfrac{{I}_{1}}{2{I}_{1}}\Rightarrow {B}_{2}=B; means, the magnetic field remains the same.

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