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Physics-

General

Easy

Question

Infinite number of straight wires each carrying current I are equally placed as shown in the figure. Adjacent wires have current in opposite direction. Net magnetic field at point P is

$\frac{μ_{0} I}{4 π} \frac{\ln 2}{\sqrt{3} a} \tilde{k}$
$\frac{μ_{0} I}{4 π} \frac{l n 4}{\sqrt{3} a} \tilde{k}$
$\frac{μ_{0} I}{4 π} \frac{l n 4}{\sqrt{3} a} (- k)$
zero

The correct answer is: $fraction numerator mu subscript 0 end subscript I over denominator 4 pi end fraction fraction numerator l n 4 over denominator square root of 3 a end fraction stack k with tilde on top$

$B equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I over denominator open parentheses acos invisible function application 30 to the power of ring operator end exponent close parentheses end fraction open parentheses sin invisible function application 30 to the power of ring operator end exponent plus sin invisible function application 30 to the power of ring operator end exponent close parentheses$
$open square brackets 1 minus fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 3 end fraction minus horizontal ellipsis.. close square brackets rightwards double arrow stack B with bar on top equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I divided by n 4 over denominator square root of 3 a end fraction stack k with bar on top$

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A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If steady current I is established in the wire as shown in the figure, the loop will

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Figure shows a square current carrying loop ABCD of side 10 cm and current i = 10 A. The magnetic moment $stack M with bar on top$ of the loop is

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A conducting rod of length l and mass m is moving down a smooth inclined plane of inclination qwith constant velocity v in fig. A current I is flowing in the conductor in a direction perpendicular to paper inwards. A vertically upward magnetic field B exists in space. Then magnitude of magnetic field B is

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The square loop ABCD, carrying a current I, is placed in a uniform magnetic field B, as shown. The loop can rotate about the axis XX. The plane of the loop makes an angle $theta open parentheses theta less than 90 to the power of ring operator end exponent close parentheses$ with the direction of B. Through what angle will the loop rotate by itself before the torque on it becomes zero–

The square loop ABCD, carrying a current I, is placed in a uniform magnetic field B, as shown. The loop can rotate about the axis XX. The plane of the loop makes an angle $theta open parentheses theta less than 90 to the power of ring operator end exponent close parentheses$ with the direction of B. Through what angle will the loop rotate by itself before the torque on it becomes zero–

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In the above reaction Z is

In the above reaction Z is

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In the given figure, $A C perpendicular C E$ and $angle A colon angle B colon angle C equals 3 colon 2 colon 1$ , then the value of $angle E C D$ is

Hence Choice 3 is correct

In the given figure, $A C perpendicular C E$ and $angle A colon angle B colon angle C equals 3 colon 2 colon 1$ , then the value of $angle E C D$ is

Hence Choice 3 is correct

Compute the value of x in the given figure.

Compute the value of x in the given figure.

parallel

Compute the value of x in the given figure.

The value of x=88^o

Compute the value of x in the given figure.

The value of x=88^o

Compute the value of x in the given figure.

Hence option B is correct

Compute the value of x in the given figure.

Hence option B is correct

Compute the value of x in the given figure.

Choice 4 is correct

Compute the value of x in the given figure.

Choice 4 is correct

parallel

In the given figure, if $P Q parallel to R S comma angle M X Q equals 135 to the power of ring operator end exponent$ and $angle M Y R equals 40 to the power of ring operator end exponent$ then the value of $angle X M Y$ is

Hence 85^o is correct option

In the given figure, if $P Q parallel to R S comma angle M X Q equals 135 to the power of ring operator end exponent$ and $angle M Y R equals 40 to the power of ring operator end exponent$ then the value of $angle X M Y$ is

Hence 85^o is correct option

In figure, if $A B parallel to C D comma C D parallel to E F$ and y : z = 3 : 7, x = ?

Hence x=126 choice 4 is correct

In figure, if $A B parallel to C D comma C D parallel to E F$ and y : z = 3 : 7, x = ?

Hence x=126 choice 4 is correct

In figure, if $A B parallel to D E comma angle B A C equals 35 to the power of ring operator end exponent$ and $angle C D E equals 53 to the power of ring operator end exponent comma angle D C E$ = ?

Hence 92^o
is correct choice

In figure, if $A B parallel to D E comma angle B A C equals 35 to the power of ring operator end exponent$ and $angle C D E equals 53 to the power of ring operator end exponent comma angle D C E$ = ?

Hence 92^o
is correct choice

parallel

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