Print all binary strings of length n

Is there any way to make this shorter or more efficient?

Yes, we can do some things.

for(int i = 0; i < Math.pow(2,n); i++)

You calculate the pow for every iteration. The pow call takes some time (compared to basic things like multiplication or addition), you should do it only once.

B = "";
B = '1'+B;
B = '0'+B;

You will create a lot of string copies for every string concatenation. This costs a lot of perfomance. In general, you should use a StringBuilder.
In this special case, we could even use a char array.

for(int i = 0; i < Math.pow(2,n); i++)
if (temp%2 == 1)
temp = temp/2;

You could use bit manipulations. But depending on the jvm or hotspot compiler, this will be done anyway.

void printB(int n)
        B = "";
        int temp = i;

Overall point: avoid abbreviations.

All together, it could be like this:

void printAllBinaryUpToLength(final int length) {
    if (length >= 63)
        throw new IllegalArgumentException("Current implementation supports only a length < 63. Given: " + length);
    final long max = 1 << length;
    for (long i = 0; i < max; i++) {
        long currentNumber = i;
        final char[] buffer = new char[length];
        int bufferPosition = buffer.length;
        while (bufferPosition > 0) {
            buffer[--bufferPosition] = (char) (48 + (currentNumber & 1));
            currentNumber >>>= 1;
        }
        System.out.println(buffer);
    }
}

Calculates (without printing, printing takes the great majority of every loop) results up to 25 (2^25 = ~33*10^6) in less than a second. Should be enough.

Here is one alternative leveraging the built-in classes

​void printB(int n)    
{         
  for(int i = 0; i < Math.pow(2,n); i++)         
  {    
    String format="%0"+n+"d";
    System.out.printf(format,Integer.valueOf(Integer.toBinaryString(i)));
    System.out.println();
  }
}