JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Magnetic Effects of Current

A current carrying rectangular loop PQRS is made of uniform wire. The length

PR = QS = 5 cm

and

PQ = RS = 100 cm

. If ammeter current reading changes from I to 2 I, the ratio of magnetic forces per unit length on the wire PQ due to wire RS in the two cases respectively

f_{PQ}^{I} : f_{PQ}^{2 I}

is :

(A) $1 : 2$
(B) $1 : 4$ ✓ Correct
(C) $1 : 5$
(D) $1 : 3$

Answer: (B)

Solution

The long sides

P Q

and

R S

, each

100

cm, are parallel wires carrying the same loop current and separated by the short side

P R = 5

cm. The force per unit length between parallel currents is

f = \frac{μ _{0} I _{1} I _{2}}{2 π d}

; since both carry the same current

I

, this gives

f \propto I^{2}

. When the reading changes from

I

2 I

the force grows by the factor

(\frac{2 I}{I})^{2} = 4

, so the ratio of forces per unit length is

f_{P Q}^{I} : f_{P Q}^{2 I} = 1 : 4

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Magnetic Effects of Current

Solution

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