JEE Main Magnetic Effect of Current PYQ

46. (AIEEE 2010 )

Two long parallel wires are at a distance $2 d$ apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field $B$ along the line $X X^{'}$ is given by

A. AIEEE 2010 Physics - Magnetic Effect of Current Question 137 English Option 1

B. AIEEE 2010 Physics - Magnetic Effect of Current Question 137 English Option 2

C. AIEEE 2010 Physics - Magnetic Effect of Current Question 137 English Option 3

D. AIEEE 2010 Physics - Magnetic Effect of Current Question 137 English Option 4

Correct Answer is Option (A)

The magnetic field varies inversely with the distance for a long conductor. That is, $B \propto \frac{1}{d} .$ According to the magnitude and direction shown graph $(1)$ is the correct one.

47. (AIEEE 2009 )

A current loop $A B C D$ is held fixed on the plane of the paper as shown in the figure. The arcs $B C$ (radius $= b$ ) and $D A$ (radius $= a$ ) of the loop are joined by two straight wires $A B$ and $C D$ . A steady current $I$ is flowing in the loop. Angle made by $A B$ and $C D$ at the origin $O$ is $30^{\circ} .$ Another straight thin wire steady current $I_{1}$ flowing out of the plane of the paper is kept at the origin. AIEEE 2009 Physics - Magnetic Effect of Current Question 139 English

The magnitude of the magnetic field $(B)$ due to the loop $A B C D$ at the origin $(O)$ is :

A. $\frac{μ_{0} I (b - a)}{24 a b}$

B. $\frac{μ_{0} I}{4 π} [\frac{b - a}{a b}]$

C. $\frac{μ_{0} I}{4 π} [2 (b - a) + π / 3 (a + b)]$

D. Zero

Correct Answer is Option (A)

The magnetic field at $O$ due to current in $D A$ is

$B_{1} = \frac{μ_{0}}{4 π} \frac{I}{a} \times \frac{π}{6}$ (directed vertically upwards)

The magnetic field at $O$ due to current in $B C$ is

$B_{2} = \frac{μ_{0}}{4 π} \frac{I}{b} \times \frac{π}{6}$ (directed vertically downwards)

The magnetic field due to current $A B$ and $C D$ at $O$ is zero.

therefore the net magnetic field is

$B = B_{1} - B_{2}$ (directed vertically upwards)

$= \frac{μ_{0}}{4 π} \frac{I}{a} \frac{π}{6} - \frac{μ_{0}}{4 π} \frac{I}{b} \times \frac{π}{6}$

$= \frac{μ_{0} I}{24} (\frac{1}{a} - \frac{1}{b})$

$= \frac{μ_{0} I}{24 a b} (b - a)$

48. (AIEEE 2008 )

A horizontal overhead powerline is at height of $4 m$ from the ground and carries a current of $100 A$ from east to west. The magnetic field directly below it on the ground is
$(μ_{0} = 4 π \times 10^{- 7} T m A^{- 1})$

A. $2.5 \times 10^{- 7} T$ southward

B. $5 \times 10^{- 6} T$ northward

C. $5 \times 10^{- 6} T$ southward

D. $2.5 \times 10^{- 7} T$ northward

Correct Answer is Option (C)

The magnetic field is

$B = \frac{μ_{0}}{4 π} \frac{2 I}{r}$

$= 10^{- 7} \times \frac{2 \times 100}{4}$

$= 5 \times 10^{- 6} T$

AIEEE 2008 Physics - Magnetic Effect of Current Question 144 English Explanation

According to right hand palm rule, the magnetic field is directed towards south.

49. (AIEEE 2007 )

Two identical conducting wires $A O B$ and $C O D$ are placed at right angles to each other. The wire $A O B$ carries an electric current $I_{1}$ and $C O D$ carries a current $I_{2}$ . The magnetic field on a point lying at a distance $d$ from $O$ , in a direction perpendicular to the plane of the wires $A O B$ and $C O D$ , will be given by

A. $\frac{μ_{0}}{2 π d} (I_{1}^{2} + I_{2}^{2})$

B. $\frac{μ_{0}}{2 π} {(\frac{I_{1} + I_{2}}{d})}^{\frac{1}{2}}$

C. $\frac{μ_{0}}{2 π d} {(I_{1}^{2} + I_{2}^{2})}^{\frac{1}{2}}$

D. $\frac{μ_{0}}{2 π d} (I_{1} + I_{2})$

Correct Answer is Option (C)

Clearly, the magnetic fields at a point $P,$ equidistant from $A O B$ and $C O D$ will have directions perpendicular to each other, as they are placed normal to each other.

$∴$ Resultant field, $B = \sqrt{B_{1}^{2} + B_{2}^{2}}$

But $B_{1} = \frac{μ_{0} I_{1}}{2 π d}$ and $B_{2} = \frac{μ_{0} I_{2}}{2 π d}$

$∴$ $B = \sqrt{{(\frac{μ_{0}}{2 π d})}^{2} (I_{1}^{2} + I_{2}^{2})}$

or, $B = \frac{μ_{0}}{2 π d} {(I_{1}^{2} + I_{2}^{2})}^{1 / 2}$

AIEEE 2007 Physics - Magnetic Effect of Current Question 145 English Explanation

50. (AIEEE 2005 )

Two concentric coils each of radius equal to $2$ $π$ $c m$ are placed at right angles to each other. $3$ ampere and $4$ ampere are the currents flowing in each coil respectively . The magnetic induction in Weber / $m^{2}$ at the center of the coils will be
$(μ = 4 π \times 10^{- 7} W b / A . m)$

A. $10^{- 5}$

B. $12 \times 10^{- 5}$

C. $7 \times 10^{- 5}$

D. $5 \times 10^{- 5}$

Correct Answer is Option (D)

AIEEE 2005 Physics - Magnetic Effect of Current Question 154 English Explanation

The magnetic field due to circular coil $1$ and $2$ are

$B_{1} = \frac{μ_{0} i_{1}}{2 r} = \frac{μ_{0} i_{1}}{2 (2 π \times 10^{- 2})}$

$= \frac{μ_{0} \times 3 \times 10^{2}}{4 π}$

$B_{2} = \frac{μ_{0} i_{2}}{2 (2 π \times 10^{- 2})} = \frac{μ_{0} \times 4 \times 10^{2}}{4 π}$

$B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{μ_{0}}{4 π} .5 \times 10^{2}$

$\Rightarrow B = 10^{- 7} \times 5 \times 10^{2}$

$\Rightarrow B = 5 \times 10^{- 5} W b / m^{2}$